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User:Kaihsu/IMlawA5

European internal market.

Section A: The scope of the four freedoms.

Chapter 5: Personal scope – public and private parties.

Notes by Kaihsu Tai, 2012-02-15.

Goods
Article 30 TFEU applies to “a transit charge designed to compensate a private undertaking for bearing costs arising from the performance by the customs and veterinary services of their tasks as providers of services in the public interest even if it has not been imposed by the State but arises from an agreement concluded by that undertaking with its customers”: Case C-16/94 Dubois.

The “State” within the meaning of Articles 34 to 36 TFEU cover not only the central government, but also local authorities (Case 45/87 Commission v Ireland (Dundalk water pipes)), the legislature and the judiciary (Case 58/80 Dansk Supermarked), public undertakings and semi-public bodies (Case 249/81 Buy Irish, where the body is financed by the State, and a Minister has the power of final decision). Bodies with statutory powers are part of the State: Case 222/82 Apple and Pear Development Council, Case 266/87 The Queen v Royal Pharmaceutical Society of Great Britain, ex parte Association of Pharmaceutical Importers.

“Statements which, by reason of their form and circumstances, give the persons to whom they are addressed the impression that they are official positions taken by the State, not personal opinions of the official, are attributable to the State”: Case C-470/03 AGM.

Failure to curb privately-created obstacles is a restriction of intra-EU trade within the meaning of Article 34 TFEU: Case C-265/95 Commission v France (Spanish strawberries), strengthened by Regulation 2679/98, confirmed in Case C-112/00 Schmidberger.

Article 34 TFEU does not have horizontal direct effect; “a contractual provision cannot be regarded as a barrier to trade [...] since it was not imposed by a Member State but agreed between individuals”: Case C-159/00 Sapod Audic, overturning Dansk Supermarked; but see Case C‑171/11 Fra.bo concerning a private-law certification body. Horizontal relations are governed by competition rules in Articles 101 and 102 TFEU.

EU institutions cannot derogate from the rules of free movement of goods, save as permitted by the Treaties: Joined Cases 80/77 and 81/77 Commissionnaires Réunis.

Natural persons
Freedom of movement of persons are provided in Articles 45 (workers), 49 (self-employed persons), and 56 (service providers and recipients) TFEU.

Relationship with citizenship rules
“Every citizen of the Union shall have the right to move and reside freely within the territory of the Member States, subject to the limitations and conditions laid down in the Treaties and by the measures adopted to give them effect”: Article 21(1) TFEU; see Directive 2004/38.

Workers
The terms ‘worker’ and ‘activity as an employed person’ define the sphere of application of one of the fundamental freedoms guaranteed by the Treaty and for that reason must be given a broad interpretation: Case 139/85 Kempf, confirmed in Joined Cases C-22/08 and C-23/08 Vatsouras. These are determined by EU law, not national law: Case 53/81 Levin.

A worker is “a person performs services of some economic value [for a certain period of time] for and under the direction of another person in return for which he receives remuneration”: Case 66/85 Lawrie-Blum, confirmed in Case C-94/07 Raccanelli. It is irrelevant that the worker is employed on a part-time basis (Levin); unable to support him or herself by his or her work without social security assistance (Kempf); or in receipt of remuneration in kind (Case 196/87 Steymann, Case C-3/87 Agegate). The motive (Levin, Steymann) and the nature of the activity (Lawrie-Blum, Case 197/86 Brown) of the worker are also irrelevant. “It is for the national court to carry out the examinations of fact necessary to determine whether that is so in the case pending before it”: Case C-456/02 Trojani.

One is not a worker if his activity “constitutes merely a means of rehabilitation or reintegration” for him: Case 344/87 Bettray. “A national of a Member State who has worked in another Member State in the context of occupational training must be regarded as a worker [...] if he has performed services in return for remuneration, provided that his activities are genuine and effective” (Case C-3/90 Bernini), rather than “on such a small scale as to be purely marginal and ancillary” (Case C-357/89 Raulin).

Status before and after employment
The freedom in Article 45 TFEU “entails the right for nationals of Member States to move freely within the territory of the other Member States and to stay there for the purposes of seeking employment”: Case C-292/89 Antonissen. Refusal “to grant a tideover allowance to a national of another Member State seeking his first employment [...] on the sole ground that he completed his secondary education in another Member State” is contrary to Article 45 TFEU: Case C-258/04 Ioannidis.

Though a jobseeker is not a worker, a Member State cannot exclude a jobseeker who have established real links with the labour market there from a jobseeker’s allowance on the basis of discriminatory criteria: Case C-138/02 Collins overruling Case 316/85 Lebon, confirmed in Vatsouras.

Former workers, who have undertaken work in another Member State and have ceased to be employed, continue to hold the status of “worker”: Case C-413/01 Ninni-Orasche. A former worker has the right “to remain in the territory of a Member State after having been employed in that State”: Article 45(3)(d) TFEU.

“A national of another Member State who undertakes university studies in the host State leading to a professional qualification, after having engaged in occupational activity in that State” has the status of “worker”: Case 39/86 Lair. A national of a Member State seeking paid employment in the territory of another Member State having worked there 17 years earlier does not have the status of “worker”: Collins.

Employers
Employers (Case C-350/96 Clean Car Autoservice) and private-sector recruiters (Case C-208/05 ITC) can invoke the rights under Article 45 TFEU “to employ [...] workers who are nationals of another Member State”, as well as invoking the limitations therein (Case C-415/93 Bosman).

Who can invoke Article 49 TFEU?: self-employed persons
The rule on equal treatment with nationals is “capable of being directly invoked by nationals of all the other member states” in matters of establishment, without further implementing measures: Case 2/74 Reyners.

Who can invoke Article 56 TFEU?: service providers and recipients
The first paragraph of Article 56 TFEU and the third paragraph of Article 57 TFEU “have direct effect and may therefore be relied on before national courts, at least in so far as they seek to abolish any discrimination against a person providing a service by reason of his nationality or of the fact that he resides in a member state other than that in which the service is to be provided”: Case 33/74 van Binsbergen.

Family members
Directive 2004/38 provides for the right of citizens of the Union and their family members to move and reside freely within the territory of the Member States. (Regulation 492/2011 provides for the freedom of movement for workers within the Union.)

“Irrespective of nationality, the family members of a Union citizen who have the right of residence or the right of permanent residence in a Member State shall be entitled to take up employment or self-employment there”: Article 23 of Directive 2004/38. In contrast, Article 11 of Regulation 1612/68 (now repealed; employment right in the host Member State for spouse and children of migrant workers) does not extend to other Member States for third-country nationals; such a person does not have the “right to take up an activity as an employed person in a Member State other than the one in which his spouse, a Community national, pursues or has pursued an activity as an employed person in exercise of her right to free movement”: Case C-10/05 Mattern and Cikotic.

“‘Family member’ means: (a) the spouse; (b) the partner with whom the Union citizen has contracted a registered partnership, on the basis of the legislation of a Member State, if the legislation of the host Member State treats registered partnerships as equivalent to marriage and in accordance with the conditions laid down in the relevant legislation of the host Member State; (c) the direct descendants who are under the age of 21 or are dependants and those of the spouse or partner as defined in point (b); (d) the dependent direct relatives in the ascending line and those of the spouse or partner as defined in point (b)”: Article 2(2) of Directive 2004/38. “Spouse” refers to a marital relationship only: Case 59/85 Reed. It is unaffected by separation: Case 267/83 Diatta, Case C-370/90 Singh. Article 13 of Directive 2004/38 provides for the retention of the right of residence by family members in the event of divorce, annulment of marriage or termination of registered partnership.

“Member States may adopt the necessary measures to refuse, terminate or withdraw any right conferred by this Directive in the case of abuse of rights or fraud, such as marriages of convenience. [...]”: Article 35 of Directive 2004/38; see also Case C-109/01 Akrich. However, the granting of “a residence permit to nationals of a non-Member State, who are members of the family of [an EU] national who has exercised his or her right of free movement” does not require the family member to “have previously been residing lawfully in another Member State”: Case C-1/05 Jia, confirmed by Case C-291/05 Eind, reinforced by C-127/08 Metock. “Without prejudice to any right to free movement and residence the persons concerned may have in their own right, the host Member State shall, in accordance with its national legislation, facilitate entry and residence for the following persons: (a) any other family members, irrespective of their nationality, not falling under the definition in point 2 of Article 2 who, in the country from which they have come, are dependants or members of the household of the Union citizen having the primary right of residence, or where serious health grounds strictly require the personal care of the family member by the Union citizen; (b) the partner with whom the Union citizen has a durable relationship, duly attested. The host Member State shall undertake an extensive examination of the personal circumstances and shall justify any denial of entry or residence to these people”: Article 3(2) of Directive 2004/38.

“The status of dependent member of a worker’s family is the result of a factual situation. The person having that status is a member of the family who is supported by the worker and there is no need to determine the reasons for recourse to the worker’s support or to raise the question whether the person concerned is able to support himself by taking up paid employment”: Lebon. Evidence of dependence “may be adduced by any appropriate means”: Jia. A mother cannot claim to be a “dependent” relative of a child only on account of emotional bonds: Case C-200/02 Zhu and Chen.

Against whom can Articles 45, 49, 56 be invoked?: horizontality
In addition to public authorities, Article 45 TFEU (workers) can be invoked against international sports federations which determine employment terms (Case 36/74 Walrave and Koch), private-law sport associations (Case C-415/93 Bosman), individual employers (Case C-281/98 Angonese), and private-law scientific research associations (Raccanelli).

Likewise, Articles 49 and 56 TFEU (establishment and services) can be invoked against sport associations with rules collectively regulating employment (Joined Cases C-51/96 and C-191/97 Deliège), private-law bodies governing an economic sector collectively (Case 90/76 van Ameyde), insurance firms with contractual conditions that disadvantage migrant workers (Case 251/83 Haug-Adrion), and trade unions (Case C-438/05 Viking Line, Case C-341/05 Laval).

Services: legal persons
Article 54 TFEU extends the right of establishment to certain companies or firms. Articles 49 and 54 TFEU “confer no right on a company incorporated under the legislation of a Member State and having its registered office there to transfer its central management and control to another Member State”: Case 81/87 Daily Mail.

“The right to form a company in accordance with the law of a Member State and to set up branches in other Member States is inherent in the exercise, in a single market, of the freedom of establishment guaranteed by the Treaty”; refusal “to register a branch of a company formed in accordance with the law of another Member State in which it has its registered office but in which it conducts no business where the branch is intended to enable the company in question to carry on its entire business in the State in which that branch is to be created” is contrary to Articles 49 and 54 TFEU: Case C-212/97 Centros, confirmed in Case C-167/01 Inspire Art.

“Where a company formed in accordance with the law of a Member State (‘A’) in which it has its registered office is deemed, under the law of another Member State (‘B’), to have moved its actual centre of administration to Member State B, [Articles 49 and 54 TFEU] preclude Member State B from denying the company legal capacity and, consequently, the capacity to bring legal proceedings before its national courts for the purpose of enforcing rights under a contract with a company established in Member State B”: Case C-208/00 Überseering.

Under Article 54 TFEU, in the absence of a uniform Community law definition of the companies which may enjoy the right of establishment on the basis of a single connecting factor determining the national law applicable to a company, the question whether Article 49 TFEU applies to a company can only be resolved by the applicable national law: Case C-210/06 Cartesio. See Regulation 2137/85 (European Economic Interest Grouping), Regulation 2157/2001 (European company statute).

Capital
Article 63(1) TFEU has direct effect: Joined Cases C-358/93 and C-416/93 Bordessa (by analogy), Joined Cases C-163/94, C-165/94 and C-250/94 Sanz de Lera (re third country), Case C-57/95 France v Commission (pension funds communication) (obiter). The main implementation is Directive 88/361.

Article 63 did not apply prior to the date of accession of a State to the EU, so it is incapable of remedying, with effect from the entry into force of the Treaty, a mortgage registration which, under the relevant national law, is vitiated from the outset by absolute and incurable nullity such as to render that registration non-existent: Case C-464/98 Stefan.

Article 63 does not allow a State to claim to be a private person in capital movement matters where the action is only possible for a public authority: Case C-478/98 Commission v Belgium (Belgian Eurobonds), Case C-98/01 Commission v UK (BAA special shares). Similarly, a national measure which derogates from general company law can violate Article 63: Joined Cases C-463/04 and C-464/04 Federconsumatori, Case C-112/05 Commission v Germany (VW blocking minority shares).

title: User:Kaihsu/IMlawA5


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User:Kaihsu/IMlawA6

European internal market.

Section A: The scope of the four freedoms.

Chapter 6: Personal scope – third-country nationals.

Notes by Kaihsu Tai, 2012-03-06.

Temporal limits and the free movement rules
A national who died before their country’s accession is not a “worker”; his family therefore cannot claim the derived rights: Case C-131/96 Mora Romero. Similarly, one unemployed at accession and has no objective possibility of obtaining work is not a “worker”: Case C-171/91 Tsiotras.

Determination of nationality
Nationality of legal persons should be available on a non-discriminatory basis: Case C-221/89 Factortame II. A Member State has to accept nationality decisions of another; dual EU/third-country nationals have EU rights: Case C-369/90 Micheletti.

Derived rights
Family members of EU citizens have derived rights: Article 7(2) of Regulation 492/2011, Article 3(2) of Directive 2004/38. These include entry (Case C-370/90 Surinder Singh) and residence (Case 267/83 Diatta) rights. To claim the rights, there must be an inter-State element, however tenuous: Case C-60/00 Carpenter. The rights remains derivative: Case C-10/05 Mattern and Cikotic. A (gratis, fast-track) visa may be required under Regulation 539/2001 or national law: Article 5(2) of Directive 2004/38.

Internal market and mobility
See:
 * Title V of Part Three TFEU (area of freedom, security and justice),
 * Joined cases 281, 283, 284, 285 and 287/85 Germany and others v Commission (migration policy),
 * Directive 2003/86 (family reunification),
 * Directive 2003/109 (long-term residents),
 * Directive 2004/114 (students),
 * Directive 2005/71 (researchers),
 * Directive 2009/50 (EU Blue Card).

An English court ruled that Article 26 TFEU (area without internal frontiers) does not confer, with direct effect, rights to third-country nationals: R v Home Secretary, ex parte Flynn [1995] 3 CMLR 397, [1997] 3 CMLR 888. Similarly see Case C-378/97 Wijsenbeek.

EU service providers can bring with them members of their workforce: Case C-113/89 Rush Portuguesa.

Member States are not required to recognize non-EU diplomas, even if recognized by another Member State: Case C-154/93 Tawil-Albertini. However, subsequent professional experience in the EU must be taken into account: Case C-319/92 Haim I.

A third-country national moving from one Member State to another and failing to exchange his driving licence cannot rely on Case C-193/94 Skanavi and Chryssanthakopoulos to evade punishment for driving without a licence: Case C-230/97 Awoyemi.

Area of freedom, security and justice
See:
 * Directive 2001/40 (expulsion of third country nationals),
 * Directive 2001/51 (financial penalties),
 * Directive 2008/115 (returning illegally staying third-country nationals),
 * Directive 2009/52 (employers of illegally staying third-country nationals),
 * Decision 2002/629/JHA (human trafficking).

Association and international agreements
In addition to the ones below, these also include cooperation agreements with Morocco, Algeria and Tunisia, as well as the Cotonou Agreement (succeeding the Lomé Convention).

Turkish Association Agreement
Decision 1/80 of the Association Council created by the Turkish Association Agreement (1963) has direct effect; the term “legal employment” therein “presupposes a stable and secure situation as a member of the labour force”: Case C-102/89 Sevince. The right to work under Article 6(1) of Decision 1/80 implies a right to reside; the limitations are to be interpreted likewise; general expulsion as a deterrent is not permissible: Case C-340/97 Nazlı. Article 10(1) (non-discrimination) of the same decision has direct effect, and precludes a Member State from excluding Turkish workers from being elected to workers chambers’ general assembly: Case C-171/01 Wählergruppe Gemeinsam. Member States cannot “adopt national legislation which excludes at the outset whole categories of Turkish migrant workers”: Case C-98/96 Ertanır.

The time limits in Article 6(1) are strictly enforced: Case C-355/93 Eroğlu. A Turkish worker’s “involuntary unemployment following the declaration of insolvency of the undertaking in which he was working cannot of itself prevent him from continuing to belong to the labour force of the host Member State”: Case C-337/07 Altun; see also Nazlı, Case C-383/03 Doğan. A Turkish worker “who has definitively ceased to belong to the labour force of a Member State because he has, for example, reached retirement age or, as in the present case, become totally and permanently incapacitated for work” do not enjoy the rights under Article 6(1): Case C-434/93 Bozkurt.

Articles 12 and 13 of the Decision 3/80 of the Association Council (social security) are not directly effective: Case C-277/94 Taflan-Met. But Article 3(1) of the same decision (equality of treatment) is directly effective: Case C-262/96 Sürül.

Derived rights of Turkish workers’ family members are provided by Article 7 of Decision 1/80. The spouse’s (conditional) right to work does not depend on a residence permit: Case C-351/95 Kadıman at 51. A stepson is a family member: Case C-275/02 Ayaz, referring to Case C-413/99 Baumbast and R; see similarly Case C-65/98 Eyüp on cohabiting ex-spouses. Rights under these decisions can be lost only by violating the limiting provisions in Article 14(1) or by leaving the host State: Case C-325/05 Derin. After vocational training, a Turkish child can respond to any employment offer “when one of his parents has in the past been legally employed in that State for at least three years”, regardless of the parent’s current residence: Case C-210/97 Akman.

Europe Agreements
Article 49 TFEU (freedom of establishment) cannot be extended to similar Europe Agreement provisions; the latter are directly effective insofar as they confer rights of entry and residence to self-employed persons; Member States can control entry and residence so far as there is no blanket prevention because of previous false declarations: Case C-63/99 Gloszczuk, Case C-235/99 Kondova, Case C-257/99 Barkoci and Malik.

The possibility that there is a disguised employment relationship not provided for by Europe Agreements does not justify systematically refusing residence: Case C-268/99 Jany.

The non-discrimination rule in Europe Agreements are directly effective: Case C-162/00 Pokrzeptowicz-Meyer, confirmed by Case C-438/00 Kolpak (horizontal direct effect for sports federations).

Requiring the worker to obtain a residence permit in advance is compatible with Europe Agreements, provided that the system for obtaining such permits is easily accessible, deals with requests in a reasonable period and objectively, and allows refusals of such permits to be contested before a court: Case C-327/02 Panayotova.

Partnership Agreements
Article 23(1) (non-discrimination in employment) of the EU–Russia Partnership Agreement (1994) is directly effective: Case C-265/03 Simutenkov (sports federation).

The bilateral EU–Switzerland Agreement on free movement of persons (1999) provides for non-discrimination; direct effect seems assumed: Case C-13/08 Stamm and Hauser; but see Case C-70/09 Hengartner and Gasser.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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User:MSB

Since 2007, I've been professor of business informatics at the Berlin School of Economics and Law. Before that I worked as an executive coach and as a manager in international companies. More specifically, I was responsible for organizing Knowledge Management for large firms like Shell and Accenture. My daughter says "you're not a real doctor" because I got my degree in physics: I cannot heal anybody but I enjoyed contributing to the creation of the World-Wide Web in the early 1990s. My little moment in the spotlight, as it were. Since 2008, I lead e-learning for the Berlin School of Economics. My main research areas are Blended Learning, especially interactive teaching in virtual environments (such as Second Life®) and lifelong education. However, I do research because I'm curious and wouldn't like to be pinned down on one topic. Besides my teaching and research, I'm also a writer of fiction.

»I'm using this Wiki as a "public science site". Public science does not only publish the results of research but leaks concepts, ideas and key points throughout the entire research process and puts them online for the purpose of open debate. All of this is 'work in progress'—please add to the growing material by editing and adding your own thoughts and opinions. That's what Wikis are there for!« (This paragraph was adapted from C. Spannagel on German Wikiversity).

Courses and Workshops (in English)

 * "Research Methods for Androids" (under development)
 * "Problem Solving 2.0: Systemic Solutions" (MBA) 2013 | 2014 | 2015
 * "E-Teaching and E-Learning with Web 2.0 Tools" (UP) WS 12
 * Int. Summer School "European Bus. & Econ. Programme" (EBEP) 2013
 * Online Identity Research: Workshop "Identity Games"
 * New project idea: Business Games
 * Virtual HWR: Practice Supervision SS 2012 | WS 2012
 * Mediawiki of the Berlin School of Economics and Law - Instructions / Documentation — Onboarding Process
 * Kurse und Veranstaltungen (Deutsche Wikiversity)

Links

 * More about me at Wikipedia
 * German Wikiversity User page
 * E-Learning blog of the Berlin School of Economics and Law.
 * Department site info
 * Video: our school's virtual islands.
 * My professional blog
 * My private (writer's) blog
 * E-Learning @ HWR Facebook page
 * Twitter
 * Vizualize. (A graphical profile using - out of date, by now - information from LinkedIn.)

de:Benutzer:Birkenkrahe

title: User:MSB


title: File:Sampling.2.B.BandPass.20120127.pdf


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User:MSB/UPotsdamPOGS11-1

= E-Learning/E-Teaching 4 POGS @ UP = This term, I'm using the Wikiversity Mediawiki for the first time to support this POGS course on E-Teaching and E-Learning for postgraduate learners at Potsdam U—inspired by the courses of my colleague C Spannagel. Though I've worked with Mediawiki before, I'm quite excited about it. Right now, you're on the Wikiversity page used to support this course. It is a subpage of the main user page, which every registered user has.

Links

 * I created a pad for my own planning. Feel free to peek and comment.
 * Wikiversity page for the German-speaking (otherwise identical) course here: check it out.
 * The Moodle course und der practice course are for registered participants only.
 * Potsdam Graduate School

Ideas — Questions — Suggestions
You can use this space to leave a comment by clicking on the "edit" link on the right (operating in complete anonymity). Any comment inserted after a " * " in the editor will appear later as a list item. Don't worry about destroying anything—it's impossible to do permanent damage here. Make sure to save your changes.

Put your own comments/ideas or suggestions for this course below: I will respond to them and you should feel free to do likewise—since this is a public venue, others can join in as well and you can tell all your friends and colleagues about it. Yes, like Facebook but without the clique-ness and without the "Like"...

Next Course:

* Monday/Tuesday March 5-6, 2012, 09.30 a.m.—5.30 p.m.   * Where: University of Potsdam, Neues Palais, Haus 8 * When: 9.30 am- 1.30 pm: room 0.75; 2.30 to 5.30 pm: room 0.69 * How to get there: map * Twitter hashtag: #ePOGS and Twitterwall

Learning by Teaching

 * Encouraging: how participants of the German course organized their own agenda around their needs on day 2 of the workshop. This way, everyone got some practice time (mostly in small, self-organized groups) on the tools of their choice: ranging from social bookmarking (delicious) over blogs (WordPress) and pads (Titanpad) to Moodle and even academic self-publishing. Important lessons from applying LdL ("Learning by Teaching")! --MSB 11:03, 15 February 2012 (UTC)

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User:Egm4313.s12.team17.deaver.md

= Reports = Report 1

= Important Links = Course Wiki Report Table Team 17 Wiki

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MediaWiki:Ipbreason

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User:Egm4313.s12.team1.wyattling/solutions

Statement
Derive the equation of motion of the spring-mass-dashpot in Fig. 53, in Kreyszig 2011 p.85, with an applied force $$r(t)$$ on the ball.

Solution
Kinematics:

$$y=y_k=y_c\!$$

Kinetics:

$$r(t)=my''+{{f}_{k}}+{{f}_{c}}\!$$

$${{f}_{k}}=k{{y}_{k}}\!$$

$${{f}_{c}}=c{{y}_{c}}'\!$$

Rearranging equations,

$$r(t)=my''+k{{y}_{k}}+c{{y}_{c}}'\!$$

$$y=y_k=y_c\!$$

Equation of motion:

$$r(t)=my''+ky+cy'\!$$

Author
Solved and typed by - --Egm4313.s12.team1.essenwein 23:06, 25 January 2012 (UTC) Reviewed By -

Solution
Kinematics, Kinetics, and Relations equations as interpreted from picture

Kinematics: $$y=y_{k}= y_{c}\!$$

Kinetics: $$f(t)=my'' + f_{k}+f_{c}\!$$

Relations:

-Force of spring relation: $$f_{k}=ky_{k}\!$$ -Force of dash-pot relation: $$f_{c}=Cy'_{c}\!$$

Putting kinematics, kinetics, and relations together:

$$y=y_{k}=y''_{c}\!$$

Using kinetic relationship $$f(t)=my''_{c}+f_{k}+f_{c}\!$$

Using the kinematics and relations equations: $$f_{k}=ky_{k}=ky_{c}\!$$ $$f(t)=my''_{c}+ky_{c}+f_{c}\!$$ Using dash-pot relation: $$f_{c}=Cy'_{c}\!$$ $$f(t)=my''_{c}+ky_{c}+Cy'_{c}\!$$

'''Final Equation

$$f(t)=my_{c}+Cy'_{c}+ky_{c}\!$$'

Problem R1.4
Given

Capacitance Equation(1):

$$Q=Cv_{c}\rightarrow \int idt=Cv_{c}\rightarrow i=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t}$$

Circuit Equation(2):

$$V=LC\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_{c}$$

First Alternate Circuit Equation (3):

$$LI''+RI'+\frac{1}{C}I=V'$$

Second Alternate Circuit Equation (4):

$$LQ''+RQ'+\frac{1}{C}Q=V$$

Statement

Using the Circuit equation (2) (see Sec2 Notes), derive equations (3) and (4) from it.

Solution

First we are asked to derive the alternate circuit equation(3) from the original circuit equation(2).

$$V=LC\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_{c}\rightarrow LI''+RI'+\frac{1}{C}I=V'$$

The first step is to take the derivative of the original equation so there is the V' term on one side:

$$V'=LC\frac{\mathrm{d^3}v_{c}}{\mathrm{d} t^3}+RC\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2}+\frac{\mathrm{d} v_{c}}{\mathrm{d} t}$$

L and C are constants and therefore remain unchanged by the derivative.

Next use the capacitance equation which states: $$I=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t}$$

Take both the first and second derivative of this equation.

First Derivative: $$I'=C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2}$$

Second Derivative: $$I''=C\frac{\mathrm{d^3} v_{c}}{\mathrm{d} t^3}$$

Now substitute in the the I terms into the appropriate places. The equation below groups terms that need to be substituted.

$$V'=L(C\frac{\mathrm{d^3}v_{c}}{\mathrm{d} t^3})+R(C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2})+\frac{({C}\frac{\mathrm{d} v_{c}}{\mathrm{d} t})}{C}$$

The final product after substitution is the alternate circuit equation:

$$LI''+RI'+\frac{1}{C}I=V'$$

We are also asked to derive the second alternate circuit equation $$LQ''+RQ'+\frac{1}{C}Q=V$$ from the original circuit equation.

Once again use the capacitance equation which also states:

$$Q=Cv_{c}\ $$

Take the first and second derivative:

First Derivative: $$Q'=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t}$$

Second Derivative: $$Q''=C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2}$$

The C term is constant and does not change when the derivative is taken.

Group the original circuit equation like in the previous problem so that parentheses mark terms that should be substituted for Q terms: $$V=L(C\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2})+R(C\frac{\mathrm{d} v_{c}}{\mathrm{d} t})+\frac{(Cv_{c})}{C}$$

After substituting, the final product is the second alternate circuit equation: $$LQ''+RQ'+\frac{1}{C}Q=V$$

Author
Solved and typed by - Chris Stewart Reviewed By -

Problem R1.5
Find a general solution for the following two problems.

Given

 * {| style="width:100%" border="0" align="left"

$$
 * $$y'' + 4y' + (\pi^2 + 4)y = 0\!$$
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the general solution.

Solution
The characteristic equation of this linear ODE with constant coefficients (L2-ODE-CC) is:

 * {| style="width:100%" border="0" align="left"|

$$
 * $$\lambda^2+4\lambda+(\pi^2+4)=0\!$$
 * $$\displaystyle (Eq. 2)

Note: The discriminant of the characteristic equation is negative.
 * }

The roots of this characteristic equation are of the form:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\lambda=-\frac{1}{2}a\pm bi$$
 * $$\displaystyle (Eq. 3)
 * }
 * }

Which yields a general solution of:

 * {| style="width:100%" border="0" align="left"

$$
 * $$y(x)=c_1e^{ax}\cos bx+c_2e^{ax}\sin bx\!$$
 * $$\displaystyle (Eq. 4)
 * }
 * }

By applying the quadratic equation to the characteristic equation, we see that the roots are:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\lambda =-2\pm (4\pi)i$$
 * $$\displaystyle (Eq. 5)
 * }
 * }

This yields a general solution of:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x)=c_1e^{-2x}\cos (4\pi*x)+c_2e^{-2x}\sin (4\pi*x)\!$$ Solved by: Egm4313.s12.team1.durrance 20:23, 25 January 2012 (UTC)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Given

 * {| style="width:100%" border="0" align="left"

$$
 * $$y''=2\pi*y'+\pi^2y\!$$
 * $$\displaystyle (Eq. 1)
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the general solution for this ODE.

Solution
This ODE is a linear ODE with constant coefficients (L2-ODE-CC).

First the equation is rearranged into general form:

 * {| style="width:100%" border="0" align="left"

$$
 * $$y''-(2\pi)y'-(\pi^2)y=0\!$$
 * $$\displaystyle (Eq. 2)
 * $$\displaystyle (Eq. 2)
 * }
 * }

The characteristic equation is:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\lambda^2-2\pi\lambda-\pi^2=0\!$$
 * $$\displaystyle (Eq. 3)
 * $$\displaystyle (Eq. 3)
 * }
 * }

Substituting the characteristic equation coefficients into the quadratic formula, we find:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\lambda_1=\lambda_2=\lambda=\pi\!$$
 * $$\displaystyle (Eq. 4)
 * $$\displaystyle (Eq. 4)
 * }
 * }

Therefore, the general form for the general solution is:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x)=c_1e^{\pi*x}+c_2xe^{\pi*x}\! $$ Solved by Egm4313.s12.team1.durrance 20:30, 25 January 2012 (UTC)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Statement
For each ODE in Fig.2 in Kreyszig 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

Given

 * {| style="width:100%" border="0" align="left"

$$
 * $$y''=g=const.\!$$
 * $$\displaystyle (Eq. 1)
 * $$\displaystyle (Eq. 1)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$
 * $$mv'=mg-bv^2\!$$
 *
$$\displaystyle (Eq. 2)
 *
$$\displaystyle (Eq. 2)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$
 * $$h'=-k\sqrt{h}\!$$
 *
$$\displaystyle (Eq. 3)
 *
$$\displaystyle (Eq. 3)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$
 * $$my''+ky=0\!$$
 *
$$\displaystyle (Eq. 4)
 *
$$\displaystyle (Eq. 4)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$
 * $$y''+w_{0}^2y=coswt, w_{0}=w\!$$
 *
$$\displaystyle (Eq. 5)
 *
$$\displaystyle (Eq. 5)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$
 * $$LI''+RI'+\frac{C}I=E'\!$$
 *
$$\displaystyle (Eq. 6)
 *
$$\displaystyle (Eq. 6)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$
 * $$EIy=f(x)\!$$
 *
$$\displaystyle (Eq. 7)
 *
$$\displaystyle (Eq. 7)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$
 * $$L\Theta''+gsin\Theta=0\!$$
 *
$$\displaystyle (Eq. 8)
 *
$$\displaystyle (Eq. 8)
 * }
 * }

Solution
In order to determine the order of any ODE, one must simply find the highest prime term in each equation.

ROUGH ANSWERS IN CASE ANYONE WANTS TO CHECK: (Okay so I think I got the superposition part, but i really want to run it past you all during our meeting before I latex everything etc.) eq. 1 = 2nd order/linear eq. 2 = 1st order/non-linear eq. 3 = 1st order/non-linear eq. 4 = 2nd order/linear eq. 5 = 2nd order/linear eq. 6 = 2nd order/linear eq. 7 = 4th order/linear eq. 8 = 2nd order/non-linear

Author
Solved and typed by - Reviewed By -


text:
User:Egm4313.s12.team20

Team Members
Rox Chang

Caitlin Sablotsky

Adam Ibrahim

Melissa Eizagaechevarria

Nicholas Calvo

Thomas Samartino

Thomas Stevens

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R1

R2

R3

R4

R5

R6

R7

title: User:Egm4313.s12.team20


text:
User:Martindolgiej

Team 16

title: User:Martindolgiej


title: File:R1.3.PNG


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User:EGM4313.s12.team6.hill

R1

title: User:EGM4313.s12.team6.hill


text:
User:Areil314/sandbox/A critique of Sample Title

(Review Paper) Cited in

sample point for
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User:Vinsiele

Iam kenyan male aged 26 years working at one of the Government law enforcing agencies ,Iam also a student at the  university of mount kenya scaling the heigths of education there,Ipersue acourse in Education biased in history.hope this page will  assist me alot in my search for knowledge.Thank you.

title: User:Vinsiele


text:
User:Egm4313.s12.team16

Members:
Martin Dolgiej

Matt Hilsenrath

Erin Standish

Brandon Wright

Nyal Jennings

Ryan McDaniel

James Welch

Reports:
R1

R2

R3

title: User:Egm4313.s12.team16


text:
User:Egm4313.s12.team16.martindolgiej

Martin Dolgiej's Page

Team 16

EGM4313 Spring 2012 Page

Commitment to Side Project

What Boredom Breeds

title: User:Egm4313.s12.team16.martindolgiej


text:
User:Egm4313.s12.team16.r1

Report 1:
By: Team 16

R1.1

From Figure: Mass, Spring, Damper system with spring and damper in parallel

First, we take the + y direction to be downward and z to be out of the page. Within this coordinate system, the displacement is described as a vector quantity y with the velocity and acceleration of the system denoted by $$ \dot{y} \! $$ and $$ \ddot{y} \! $$ respectively.

Multiplying by the mass denoted $$ m \! $$ we derive the kinematics of the system to be

$$ m\ddot{y}+F1=F_{y} \! $$ $$ F1=F_{k}+F_{c} \! $$ To develop the kinetics of the system, we consider the two forces:

For the force limiting displacement in the y direction, we look at Hooke's law

$$ F_{k}=-ky \! $$

For the force opposing the motion, the damping is related to the velocity of the system by

$$ F_{c}=-c\dot{y} \! $$

where $$ c \! $$ is the coefficient of viscous friction

Equating the kinematics to the kinetics we preliminarily have

$$ m\ddot{y}+ky+c\ddot{y}=\ddot{F_{y}} \! $$ and simplifying with: $$ y=y_{k}=y_{c} \! $$ $$ \dot{y}=\dot{y_{c}}=\dot{y_{k}} \! $$

Final Equation:

Author: Nyal Jennings

Last Edit: 01/28/2012 - 11:01 PM

Reviewed By: Martin Dolgiej

R1.2
From Figure 1 - Spring:

Mass:

Damp:

Forcing Funtion:

Balancing the forces in the y direction: $$ \sum F_y = F_m - F_r = F_k - F_c \! $$

Substituting equations (1),(2),(3), and (4): $$ my'' - r(t) = -ky - cy' \! $$

Author: Martin Dolgiej

Last Edit: 01/28/2012 - 11:32 PM

Reviewed By: Matt Hilsenrath

R1.3
FBD of Spring

$$f_s = k*y_s \! $$

FBD of Dashpot

$$ Cy_D' = f_c \! $$

FBD of Mass

(Newton's 2nd Law)

$$ f = my'' \! $$

$$\sum F(t) = my'' + f_s - f_c \! $$

$$ f_1 = f_s - F_c \! $$

Author: Ryan McDaniel

Reviewed by: Erin Standish

R1.4
Given:

Solution:

Deriving equation (3):

Taking the derivative of (Eq. 2) gives:

By definition:

$$ i=C\frac{dv_C}{dt} $$

$$ i'=C\frac{d^2v_C}{dt^2} $$

$$ i''=C\frac{d^3v_C}{dt^3} $$

Plugging theses definitions into (Eq. A):

Deriving equation (4):

By definition:

$$ Q=Cv_c \! $$

$$ Q'=C\frac{dv_C}{dt} $$

$$ Q''=C\frac{d^2v_C}{dt^2} $$

Using Q and its time derivatives, (2) can be written as

Authors: Erin Standish, James Welch

Reviewed by: Brandon Wright Ryan McDaniel

R1.5
For question 1.5a we have to find the general solution to the equation: (1) and we have the relation (2) Its first two derivatives are (3) and (4) Incorporating (2), (3), and (4) into equation (1), it can be shown that Factoring out the exponential term, We recognize that an exponential term cannot equal zero, so it can be divided out. Now we are left with the characteristic equation Recognizing the relation, (7) can be factored to to reveal the roots of the characteristic equation to be and so the standard form of the equation can be written For question 1.5b we have to find the general solution to the equation: and we have the relation its first two derivatives are and Incorporating these equations it can be shown that Dividing out, we have and since can't equal 0, we must evaluate the roots of the equation. Factoring we have, Which places a double root at So to describe the system in standard form,

Author: Brandon Wright Reviewed by: Matt Hilsenrath

Part A:
Order:     Second Order ODE Linearity: Linear Superposition: $$ \bar{y}(x) = y_p(x) + y_h(x) \! $$ From Figure 1 - $$ y'' = g \! $$ Splitting differential equation into particular and homogeneous: $$ y_p'' = g \! $$ $$ y_h'' = 0 \! $$ Summing above equations together: $$ y_p + y_h = g \! $$ Therefore superposition IS valid.

Part B:
Order:     First Order ODE Linearity: Non-linear Superposition: $$ \bar{v}(x) = v_p(x) + v_h(x) \! $$ From Figure 2 - $$ mv' = mg - bv^2 \! $$ Splitting differential equation into particular and homogeneous: $$ mv_p' + bv_p^2 = mg \! $$ $$ mv_h' + bv_h^2 = 0 \! $$ Summing above equations together: $$ mv_p' + mv_h' + bv_p^2 + bv_h^2 = mg \! $$ Therefore, superposition is NOT valid.

Part C:
Order:     First Order ODE Linearity: Non-linear Superposition: $$ \bar{h}(x) = h_p(x) + h_h(x) \! $$ From Figure 3 - $$ h' = -kh^{1/2} \! $$ Splitting differential equation into particular and homogeneous: $$ h_h' = -kh_h^{1/2} \! $$ $$ h_p' = -kh_p^{1/2} \! $$ Summing above equations together: $$ h_h' + h_p' = -kh_h^{1/2} - kh_p^{1/2} \! $$ Therefore superposition is NOT valid.

Part D:
Order:     Second Order ODE Linearity: Linear Superposition: $$ \bar{y}(x) = y_p(x) + y_h(x) \! $$ From Figure 4 - $$ my'' + ky = 0 \! $$ Splitting differential equation into particular and homogeneous: $$ my_h'' + ky_h = 0 \! $$ $$ my_p'' + ky_p = 0 \! $$ Summing above equations together: $$ my_h + my_p + ky_h + ky_p = 0 \! $$ Therefore superposition IS valid.

Part E:
Order:     Second Order ODE Linearity: Linear Superposition: $$ \bar{y}(x) = y_p(x) + y_h(x) \! $$ From Figure 5 - $$ y'' + w_0^2y = \cos{wt} \! $$ Splitting differential equation into particular and homogeneous: $$ y_h'' + w_0^2y_h = 0 \! $$ $$ y_p'' + w_0^2y_p = \cos{wt} \! $$ Summing above equations together: $$ y_p + y_h + w_0^2y_p + w_0^2y_h = \cos{wt} \! $$ Therefore superposition IS valid.

Part F:
Order:     Second Order ODE Linearity: Linear Superposition: $$ \bar{I}(x) = I_p + I_h \! $$ From Figure 6 - $$ LI'' + RI' + \frac{1}{C}I = E' \! $$ Splitting differential equation into particular and homogeneous: $$ LI_p'' + RI_p' + \frac{1}{C}I_p = E' \! $$ $$ LI_h'' + RI_h' + \frac{1}{C}I_h = 0 \! $$ Summing above equations together: $$ LI_p + LI_h + RI_p' + RI_h' + \frac{1}{C}I_p + \frac{1}{C}I_h = E' \! $$ Therefore superposition IS valid.

Part G:
Order:     iv Order ODE Linearity: Non-linear Superposition: $$ \bar{y}(x) = y_p + y_h \! $$ From Figure 7 - $$ EIy^{iv} = f(x) \! $$ Splitting differential equation into particular and homogeneous: $$ EIy_p^{iv} = f(x) \! $$ $$ EIy_h^{iv} = 0 \! $$ Summing above equations together: $$ EIy_p^{iv} + EIy_h^{iv} = f(x) \! $$ Therefore superposition is NOT valid.

Part H:
Order:     Second Order ODE Linearity: Non-linear Superposition: $$ \bar{\theta} = \theta_p + \theta_h \! $$ From Figure 8 - $$ L\theta'' + g\sin{\theta} = 0 \! $$ Splitting differential equation into particular and homogeneous: $$ L\theta_h'' + g\sin{\theta_h} = 0 \! $$ $$ L\theta_p'' + g\sin{\theta_p} = 0 \! $$ Summing above equations together: $$ L\theta_h + L\theta_p + g\sin{\theta_h} + g\sin{\theta_p} = 0 \! $$ Therefore superposition is NOT valid.

Author: Martin Dolgiej Last Edited: 01/28/2011 - 4:10 PM Reviewed by:James Welch, Matt Hilsenrath

title: User:Egm4313.s12.team16.r1


text:
University of Florida/Egm4313/s12.team6.hill/R1

Problem 1: Spring-dashpot system in parallel
Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$ f(t) $$

Given
Spring-dashpot system in parallel

Solution
The kinematics of the system can be described as,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle x = x_k = x_c $$
 *  $$\longrightarrow(1) $$
 * }
 * }

The kinetics of the system can be described as,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle m\ddot{x} + f_I = f(t) $$ and,
 *  $$\longrightarrow(2) $$
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_I = f_k + f_c $$
 *  $$\longrightarrow(3) $$
 * }
 * }

Given that,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_k = kx_k $$
 *  $$\longrightarrow(3a) $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_c = c\dot{x_c} $$
 *  $$\longrightarrow(3b) $$
 * }
 * }

From (1), it can be found that,
 * {| style="width:100%" border="0" align="left"

\dot{x} = \dot{x_k} = \dot{x_c} $$ and,
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\ddot{x} = \ddot{x_k} = \ddot{x_c} $$
 * $$\displaystyle
 * }

From (3), it can be found that,
 * {| style="width:100%" border="0" align="left"

f_I =c\dot{x_c} + kx_k $$
 * $$\displaystyle
 * }

Finally, it can be found that
 * {| style="width:100%" border="0" align="left"

$$\displaystyle m\ddot{x} + c\dot{x} + kx = f(t) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }


text:
User:Egm4213.s12.team7.melroy.j

Jennifer Melroy

title: User:Egm4213.s12.team7.melroy.j


text:
User:Egm4313.s12.team20/R1

Statement
Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$\displaystyle f(t). $$ The spring-dashpot system is shown in Figure 1.

Figure 1

Solution
Kinematics:
 * {| style="width:100%" border="0"|-

y = y_{k} = y_c$$
 * style="width:95%" | $$ \displaystyle
 * (1.0)
 * }

Kinetics:
 * {| style="width:100%" border="0"|-

f(t) = my'' + f_1 $$
 * style="width:95%" | $$ \displaystyle
 * (1.1)
 * }

 * {| style="width:100%" border="0"|-

 * style="width:95%" | || (1.2)
 * }

By Substitution of $$ \displaystyle f_1 $$ into equation (1.1) we obtain
 * {| style="width:100%" border="0"|-

f(t) = my'' + f_k + f_c $$ Constitutive Relations: -Force of Spring
 * style="width:95%" | $$ \displaystyle
 * (1.3)
 * }
 * {| style="width:100%" border="0"|-

f_k = ky_k $$
 * style="width:95%" | $$ \displaystyle
 * (1.4)
 * }

-Force of Dash-Pot
 * {| style="width:100%" border="0"|-

f_c = cy'_c$$ Because of the kinematic relations where $$ \displaystyle y_c = y_k $$ we manipulate equation (1.4) and obtain
 * style="width:95%" | $$\displaystyle
 * (1.5)
 * }
 * {| style="width:100%" border="0"|-

f_k = ky_k = ky_c$$
 * style="width:95%" | $$\displaystyle
 * (1.6)
 * }

Therefore, by subsitution into equation (1.2)
 * {| style="width:100%" border="0"|-

 * style="width:95%" | || (1.7)
 * }

After finally substituting equation (1.7) into the original f(t) equation (1.1), the final equation equals

Given
Figure 2

Solution
Kinematics:
 * {| style="width:100%" border="0"|-

y=y_k=y_c$$
 * style="width:95%" | $$ \displaystyle
 * (2.0)
 * }

Kinetics:

 * {| style="width:100%" border="0"|-

r(t)=-my''-f_k-f_c$$
 * style="width:95%" | $$ \displaystyle
 * (2.1)
 * }

 * {| style="width:100%" border="0"|-

f_k=ky_k$$
 * style="width:95%" | $$ \displaystyle
 * (2.2)
 * }

 * {| style="width:100%" border="0"|-

f_c=cy_c'$$
 * style="width:95%" | $$ \displaystyle
 * (2.3)
 * }

Given
Free body diagram:

Solution
By definition:

$$\displaystyle a=y'' $$

(3.1)

$$\displaystyle f_k=ky_k$$ (3.2)

$$\displaystyle f_c=cy_c' $$ (3.3)

Since the spring, dashpot and mass are aligned, the forces on the dashpot are equal and opposite and can be described as the internal forces.

$$ \displaystyle f_k=f_c=f_I $$ (3.4)

The net force of the system is the difference between the applied force and the internal force.

$$ \displaystyle F_s=f(t)-f_I $$ (3.5)

Applying Newtons 2nd law:

$$\displaystyle F_s=m*a $$  (3.6)

Plug in equation (3.1) and (3.5) into equation (3.6):

$$\displaystyle

my''=f(t)-f_I $$  (3.7)

Rearrange:

Given
Circuit Equation:
 * {| style="width:100%" border="0"

 * style="width:95%" | $$ V = LC\frac{d^2v_{c}}{dt^2} + RC\frac{dv_{c}}{dt} + v_{c}$$
 * (4.0)
 * }

Capacitance:
 * {| style="width:100%" border="0"|-

 * style="width:95%" | $$ Q=Cv_{c} \rightarrow \int idt=Cv_{c} \rightarrow i = C\frac{dv_{c}}{dt}$$
 * (4.1)
 * }

Solution
Alternate Circuit Equation 1

Current Equation:
 * {| style="width:100%" border="0"

 * style="width:95%" | $$ I = C\frac{dv_{c}}{dt}$$
 *
(4.2)
 * }

Derive Equation 4.2:
 * {| style="width:100%" border="0"|-

 * style="width:95%" | $$ I' = C\frac{d^2v_{c}}{dt^2}$$
 *
(4.3)
 * }

Rearrange the Current Intergral from Equation 4.1:
 * {| style="width:100%" border="0"|-

 * style="width:95%" | $$ \int idt=Cv_{c} \rightarrow v_{c}=\frac{1}{C}\int idt $$
 *
(4.4)
 * }

Substitute Equations 4.2, 4.3 and 4.4 into Equation 4.0:
 * {| style="width:100%" border="0"|-

$$ \Downarrow $$
 * style="width:95%" | $$ V = LC\frac{d^2v_{c}}{dt^2} + RC\frac{dv_{c}}{dt} + v_{c}$$
 * }
 * {| style="width:100%" border="0"|-
 * {| style="width:100%" border="0"|-

 * style="width:95%" | $$ V = LI'+ RI + \frac{1}{C}\int idt $$
 *
(4.5)
 * }

Derive Equation 4.5 to find Alternate Circuit Equation 1:
 * {| style="width:100%" border="0"|-

$$ \Downarrow $$
 * style="width:95%" | $$ V = LI'+ RI + \frac{1}{C}\int idt $$
 * }
 * }

Alternate Circuit Equation 2

Charge Equation:
 * {| style="width:100%" border="0"

 * style="width:95%" | $$ Q = Cv_{c} \ $$
 *
(4.7)
 * }

Derive Equation 4.7:
 * {| style="width:100%" border="0"|-

 * style="width:95%" | $$ Q'=C\frac{dv_{c}}{dt}$$
 *
(4.8)
 * }

Derive Equation 4.8:
 * {| style="width:100%" border="0"|-

 * style="width:95%" | $$ Q''=C\frac{d^2v_{c}}{dt^2}$$
 *
(4.9)
 * }

Rearrange the Charge equation from Equation 4.1:
 * {| style="width:100%" border="0"|-

 * style="width:95%" | $$ Q = Cv_{c} \rightarrow v_{c}=\frac{Q}{C}$$
 *
(4.10)
 * }

Substitute Equations 4.8, 4.9 and 4.10 into Equation 4.0:
 * {| style="width:100%" border="0"|-

$$ \Downarrow $$
 * style="width:95%" | $$ V = LC\frac{d^2v_{c}}{dt^2} + RC\frac{dv_{c}}{dt} + v_{c}$$
 * }
 * }

R 1.5
Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.4-5

Problem 4
Find a general solution. Check your answer by substitution.

Given

 * {| style="width:100%" border="0"

$$\displaystyle y''+4y'+(\pi^2 + 4)y = 0$$ (5.0)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
Rewrite as

 * {| style="width:100%" border="0"

$$\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y=0 $$ (5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Set value for lambda
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$ (5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Plug in the variable lambda to get the quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+4\lambda+(\pi^2+4)=0$$ (5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The formula for the Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha \pm \beta i$$ (5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the quadratic formula, the roots are determined to be
 * {| style="width:100%" border="0"

$$\lambda =-2\pm (\pi)i$$ (5.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The General Solution formula for Complex Conjugate roots is
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$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1\cos \beta x+c_2\sin \beta x)$$ (5.6)
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Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Problem 5
Find a general solution. Check your answer by substitution.

Given

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$$\displaystyle y''+2\pi y'+\pi^2y = 0$$ (5.8)
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Solution
Rewrite as

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$$\frac{d^2 y}{dx^2} + 2\pi \frac{dy}{dx} + (\pi^2 y)=0 $$ (5.9)
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Set value for lambda (refer to equation 5.2)
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$$\lambda = \frac{dy}{dx}$$
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Plug in the variable lambda to get the quadratic equation
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$$\displaystyle \lambda^2+2\pi \lambda+\pi^2 y=0$$ (5.10)
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The formula for the Real Double root is
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$$\lambda=-\frac{1}{2}\alpha$$ (5.11)
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Using the quadratic formula, the root is determined to be
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$$\displaystyle\lambda=-\pi$$ (5.12)
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The General Solution formula for Real Double root is
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$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1 + c_2 x)$$ (5.13)
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Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Given
For each ODE in Fig.2 in Kreyszig 2011 p.3 (except the last one involving a system of 2 ODE's), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied:

Falling stone
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$$  \displaystyle y''=g=const. $$     (6.0)
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Parachutist
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$$  \displaystyle mv'=mg-bv^2 $$     (6.1)
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Outflowing water
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$$  \displaystyle h'=-k\sqrt{h} $$     (6.2)
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Vibrating mass on a spring
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$$  \displaystyle my''+ky=0 $$     (6.3)
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Beats of a vibrating system
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$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0}\approx w $$ (6.4)
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Current I in an RLC circuit
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$$  \displaystyle LI''+RI'+\frac{C}I=E' $$     (6.5)
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Deformation of a beam
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$$  \displaystyle EIy^{iv}=f(x) $$     (6.6)
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Pendulum
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$$  \displaystyle L\Theta''+g\sin\Theta=0 $$     (6.7)
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Solution
The order of an ODE is determined by the order of the highest-order derivatives present in the equation.

A linear ODE is one in which the dependent variable, y in this case, and its derivatives in additive combinations of their first powers with coefficients that are constants or that depend on an independent variable. If the equation does not fit the above specifications, the equation is said to be nonlinear.

The superposition principle states that if $$y_p$$ (particular) and $$y_h$$ (homogeneous) are both solutions to a differential equation:

$$ \displaystyle ay_p'' + by_p' + cy_p=f_p(x)$$

$$ \displaystyle ay_h'' + by_h' + cy_h=f_h(x)$$

Then when the two equations are added together, substituting in:

$$ \displaystyle y=\bar{y} = y_p + y_h $$

You will obtain the original ODE.

6.0

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$$  \displaystyle y''=g=const. $$     (6.0)     Order: $$2^{nd}$$
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Linearity: Linear

Superposition: Yes

Examining both particular and homogeneous solutions: Particular solution: $$ \displaystyle y_p'' = g$$

Homogeneous solutions: $$ \displaystyle y_{h1}'' = 0$$

$$ \displaystyle y_{h2}'' = 0$$

Adding the two homogeneous solutions together, results in the following: $$ \displaystyle y_h = (y_{h1} + y_{h2}) = 0$$

Let: $$ \displaystyle \bar{y} = y_h + y_p''$$

Adding $$y_p$$ and $$y_h$$ equations together results in: $$ \displaystyle (y_p + y_h) = g $$

Which simplifies to: $$ \displaystyle \bar{y}'' = g $$

Thus, the superposition principle can be applied.

6.1

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$$  \displaystyle mv'=mg-bv^2 $$     (6.1)     Order: $$1^{st}$$
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Linearity: Nonlinear

Superposition: No

Homogeneous solution: $$ \displaystyle mv_h' + bv_h^2 = 0 $$
 (6.1.1)

Particular solution: $$ \displaystyle mv_p' + bv_p^2 = mg $$
 (6.1.2)

Let $$ \displaystyle \bar{v} = v_h + v_p $$

Adding (6.1.1) and (6.1.2): $$ \displaystyle m(v_h' + v_p') + b(v_h^2 + v_p^2) = mg $$

Likewise $$ \displaystyle m\bar{v} + b\bar{v}^2 = mg $$

However $$ \displaystyle \bar{v}^2 = (v_h + v_p)^2 \not= (v_h^2 + v_p^2) $$

Thus $$ \displaystyle \bar{v} $$ is not a solution and superposition does not apply.

6.2

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$$  \displaystyle h'=-k\sqrt{h} $$     (6.2)     Order: $$1^{st}$$
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Linearity: Nonlinear

Superposition: No

Homogeneous solution: $$ \displaystyle h_{h}' + k\sqrt{h_{h}} = 0 $$

Particular solution: $$ \displaystyle h_{p}' + k\sqrt{h_{p}} = 0 $$

Adding $$h_p'$$ and $$h_h'$$ equations together results in: $$ \displaystyle h_{h}' + k\sqrt{h_{h}} + h_{p}' + k\sqrt{h_{p}} = 0 $$

Which can be rearranged to: $$ \displaystyle (h_{h}' + h_{p}') + k(\sqrt{h_{h}} + \sqrt{h_{p}}) = 0 $$

Simplify using: $$ \displaystyle \bar{h} = h_{p} + h_{h} $$

To: $$ \displaystyle \bar{h'} + k(\sqrt{h_{h}} + \sqrt{h_{p}}) = 0 $$

Since: $$ \displaystyle k(\sqrt{h_{h}} + \sqrt{h_{p}}) \not= k\sqrt{h_{h} + h_{p}} $$

Then: $$ \displaystyle \bar{h'} + k(\sqrt{h_{h}} + \sqrt{h_{p}}) \not= \bar{h'} + k\sqrt{\bar{h}} $$

So superposition can not be applied.

6.3

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$$  \displaystyle my''+ky=0 $$     (6.3)     Order: $$2^{nd}$$
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Linearity: Linear

Superposition: Yes

$$ \displaystyle y_1'' + \frac{k}{m}y_1 = 0 $$
 (6.3.1)

$$ \displaystyle y_2'' + \frac{k}{m}y_2 = 0 $$
 (6.3.2)

$$ \displaystyle c_1 * (6.3.1) + c_2 * (6.3.2) = (c_1y_1 + c_2y_2)'' + \frac{k}{m}(c_1y_1 + c_2y_2) = 0 $$

Therefore, $$ \displaystyle y = c_1y_1 + c_2y_2 $$

Thus the principle of superposition applies.

6.4

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$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0} \approx w $$ (6.4)    Order: $$2^{nd}$$
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Linearity: Linear

Superposition: Yes

Homogeneous solution: $$ \displaystyle y_h'' + w_{0}^2 y_h = 0 $$

Particular solution: $$ \displaystyle y_p'' + w_{0}^2 y_p = \cos{wt} $$

Examining homogeneous solution: $$ \displaystyle y_{h1}'' + w_{0}^2y_{h1} = 0 $$
 (6.4.1)

$$ \displaystyle y_{h2}'' + w_{0}^2y_{h2} = 0 $$
 (6.4.2)

$$ \displaystyle c_1 * (6.4.1) + c_2 * (6.4.2) = (c_1y_{h1} + c_2y_{h2})'' + w_{0}^2(c_1y_{h1} + c_2y_{h2}) = 0 $$

Therefore $$ \displaystyle y_h = c_1y_{h1} + c_2y_{h2} $$

And also $$ \displaystyle y(t) = y_h + y_p = c_1y_{h1} + c_2y_{h2} + y_p $$

Superposition is valid.

6.5

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$$  \displaystyle LI''+RI'+\frac{C}I=E' $$     (6.5)     Order: $$2^{nd}$$
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Linearity: Linear

Superposition: Yes

Homogeneous solution: $$ \displaystyle LI_h'' + RI_h' + (1/C)I_h = 0 $$

Particular solution: $$ \displaystyle LI_p'' + RI_p' + (1/C)I_p = E' $$

Examining the homogeneous solution: $$ \displaystyle LI_{h1}'' + RI_{h1}' + (1/C)I_{h1} = 0 $$
 (6.5.1)

$$ \displaystyle LI_{h2}'' + RI_{h2}' + (1/C)I_{h2} = 0 $$
 (6.5.2)

$$ \displaystyle c_1 * (6.5.1) + c_2 * (6.5.2) = L(c_1I_{h1} + c_2I_{h2})'' + R(c_1I_{h1} + c_2I_{h2})' + (1/C)(c_1I_{h1} + c_2I_{h2})$$

Therefore $$ \displaystyle I_h = c_1I_{h1} + c_2I_{h2} $$

And $$ \displaystyle I = I_h + I_p = c_1I_{h1} + c_2I_{h2} + I_p $$

Superposition is valid.

6.6

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$$  \displaystyle EIy^{iv}=f(x) $$     (6.6)     Order: $$4^{th}$$
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Linearity: Linear

Superposition: Yes

Homogenous solution: $$ \displaystyle EIy_h^{iv} = 0 $$

Particular solution: $$ \displaystyle EIy_p^{iv} = f(x) $$

Examining the homogeneous solution:

$$ \displaystyle EIy_{h1}^{iv} = 0 $$
 (6.6.1)

$$ \displaystyle EIy_{h2}^{iv} = 0 $$
 (6.6.2)

$$ \displaystyle EIy_{h3}^{iv} = 0 $$
 (6.6.3)

$$ \displaystyle EIy_{h4}^{iv} = 0 $$
 (6.6.4)

$$ \displaystyle c_1 * (6.6.1) + c_2 * (6.6.2) + c_3 * (6.6.3) + c_4 * (6.6.4) = EI(y_{h1} + y_{h2} + y_{h3} + y_{h4} )^{iv} = 0 $$

Therefore $$ \displaystyle y_h = c_1y_{h1} + c_2y_{h2} + c_3y_{h3} + c_4y_{h4} $$

Also $$ \displaystyle y(x) = y_h + y_p = c_1y_{h1} + c_2y_{h2} + c_3y_{h3} + c_4y_{h4} + y_p $$

Superposition applies.

6.7

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$$  \displaystyle L\Theta''+g\sin\Theta=0 $$     (6.7)
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Order: $$2^{nd}$$

Linearity: Nonlinear

Superposition: No

Let $$ \displaystyle \Theta_1 $$ and $$ \displaystyle \Theta_2 $$ be solutions to the homogeneous ODE above: $$ \displaystyle L\Theta_1'' + g\sin\Theta_1 = 0 $$
 (6.7.1)

$$ \displaystyle L\Theta_2'' + g\sin\Theta_2 = 0 $$
 (6.7.2)

For superposition: Let $$ \displaystyle \bar\Theta = c_1\Theta_1 + c_2\Theta_2 $$

$$ \displaystyle c_1 * (6.7.1) + c_2 * (6.7.2) = L(c_1\Theta_1 + c_2\Theta_2)'' + g(c_1\sin\Theta_1 + c_2\sin\Theta_2) = 0 $$

$$ \displaystyle L\bar\Theta'' + g\sin\bar\Theta = 0 $$

Comparing: $$ \displaystyle (c_1\Theta_1 + c_2\Theta_2) = \bar\Theta $$ BUT $$ \displaystyle \sin\bar\Theta \not= c_1\sin\Theta_1 + c_2\sin\Theta_2 $$

Therefore $$ \displaystyle \bar\Theta $$ is not a solution and superposition cannot be used.

title: User:Egm4313.s12.team20/R1


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User:Egm4313.s12.team13.steinberg

R1

R2

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User:Egm4313.s12.team16.brandonwright

Yo

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User:Egm4313.s12.team16.jameswelch

Team site

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User:Egm4313.s12.team1.essenwein

Eric Essenwein

Links
Team Page

Course Wiki

Statement
Consider the L2-ODE-CC (5) p7b-7 with the window function f(x) p.9-8 as excitation:
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$$\displaystyle{y''-3y'+2y=r(x)}$$ $$r(x)=f(x)$$ (4.0) and the initial conditions
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$$\displaystyle{y(0)=1, y'(0)=0}$$ (4.1) 1. Find $$ y_n(x) $$ such that:
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$$\displaystyle{y''_n +ay'_n + by_n=r_n(x)}$$ (4.2) with the same initial conditions (4.1). Plot $$ y_n(x)$$ for n=2,4,8, for x in [0,10]. 2.Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution. Level 1:n=0,1
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Part 1 Solution
First, we shift the excitation f(x) to the left by introducing a new independent variable, t. This allows the period to start at zero.
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$$\displaystyle t=x-\frac{1}{4}$$ (4.3) The piecewise representation of the window function is now (in terms of t) as follows:
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$$\displaystyle f(t)=A$$ for $$\displaystyle t=\left[ 0,2 \right]$$ $$\displaystyle f(t)=0$$ for $$\displaystyle t=\left[ 2,4 \right]$$ (4.4) To find the Fourier transform, the period of oscillation is determined:
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$$\displaystyle p=2L=4$$ (4.5) And the frequency of oscillation:
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$$\displaystyle \omega =\frac{2\pi }{p}=\frac{\pi }{2}$$ (4.6) The general form of a Fourier transform is
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$$\displaystyle f(t)={{a}_{0}}+\sum\limits_{1}^{n}{\left[ {{a}_{n}}\cos (n\omega t)+{{b}_{n}}\sin (n\omega t) \right]}$$ (4.7) The following equations are given values of the constants in (4.7), evaluated for the function in (4.4):
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$$\displaystyle {{a}_{0}}=\frac{1}{2L}\int\limits_{0}^{2L}{f(t)dt}=\frac{1}{4}\int\limits_{0}^{2}{Adt}=\frac{A}{4}(2-0)=\frac{A}{2}$$ (4.8)
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$$\displaystyle {{a}_{n}}=\frac{1}{L}\int\limits_{0}^{2L}{f(t)\cos (n\omega t)dt} =\frac{1}{2}\int\limits_{0}^{2}{A\cos (\frac{n\pi }{2}t)dt} =\frac{A}{n\pi }(\sin n\pi -\sin 0)$$ (4.9)
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$$\displaystyle {{b}_{n}}=\frac{1}{L}\int\limits_{0}^{2L}{f(t)\sin (n\omega t)dt}=\frac{1}{2}\int\limits_{0}^{2}{A\sin (\frac{n\pi }{2}t)dt}=-\frac{A}{n\pi }(\cos n\pi -\cos 0)=\frac{A}{n\pi }(1-\cos n\pi )$$ (4.10) In their simplest forms:
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$$\displaystyle {{a}_{n}}=0$$ (4.11)
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$$\displaystyle {{b}_{n}}=\left\{ 0,\frac{A}{\pi },0,\frac{A}{2\pi },0,\frac{A}{3\pi },0,\frac{A}{4\pi },...,0,\frac{2A}{n\pi } \right\}$$ (4.12) Plugging in r(x)=f(t) in (4.0),
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$$\displaystyle y''-3y'+2y=\frac{1}{2}+\frac{2}{n\pi }\sin (\frac{n\pi }{2}t)$$ (4.13) From the general form, a particular solution to (4.13) and its derivatives are as follows:
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$$\displaystyle {{y}_{n}}={{A}_{n}}\cos (\frac{n\pi }{2}t)+{{B}_{n}}\sin (\frac{n\pi }{2}t)$$ (4.14)
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$$\displaystyle {{y}_{n}}'=-{{A}_{n}}\frac{n\pi }{2}\sin (\frac{n\pi }{2}t)+{{B}_{n}}\frac{n\pi }{2}\cos (\frac{n\pi }{2}t)$$ (4.15)
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$$\displaystyle {{y}_{n}}''=-{{A}_{n}}\frac{4}\cos (\frac{n\pi }{2}t)-{{B}_{n}}\frac{4}\sin (\frac{n\pi }{2}t)$$ (4.16)
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By plugging in (4.14), (4.15), and (4.16) into (4.13), we find An and Bn in terms of another constant, Cn:

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$$\displaystyle {{A}_{n}}=6n\pi {{C}_{n}}$$ (4.17)
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$$\displaystyle {{B}_{n}}=(8-{{n}^{2}}{{\pi }^{2}}){{C}_{n}}$$ (4.18)
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$$\displaystyle {{C}_{n}}={{\left[ \tfrac{1}{8}n\pi ({{(8-{{n}^{2}}{{\pi }^{2}})}^{2}}+{{(6n\pi )}^{2}}) \right]}^{-1}}$$ (4.19) After substituting t with (4.3), the solutions are shown for n=2,4,8.
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$$\displaystyle {{y}_{2}}={{y}_{1}}+{{y}_{2}}$$ (4.20)
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$$\displaystyle {{y}_{4}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}$$ (4.21)
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$$\displaystyle {{y}_{8}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}+{{y}_{5}}+{{y}_{6}}+{{y}_{7}}+{{y}_{8}}$$ (4.22)
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Part 2 Solution
The excitation for n=0 and n=1 are the same, because $$a_1=b_1=0$$.

The homogeneous solution to (4.0) is
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$$\displaystyle {{y}_{h}}={{C}_{1}}{{\operatorname{e}}^{2t}}+{{C}_{2}}{{\operatorname{e}}^{t}}$$ (4.23) For $$r(x)=a_0$$, the particular solution is
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$$\displaystyle {{y}_{p}}={{C}_{3}}$$ (4.24) Evaluating, we find that
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$${{C}_{1}}=-\tfrac{3}{4}$$ $${{C}_{2}}=\tfrac{3}{2}$$ $${{C}_{3}}=\tfrac{1}{4}$$ (4.25) Thus, the complete solution $$y_0$$ is
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$$ {{y}_{0}}=\tfrac{1}{4}-\tfrac{3}{4}{{\operatorname{e}}^{2(x-\tfrac{1}{4})}}+\tfrac{3}{2}{{\operatorname{e}}^{x-\tfrac{1}{4}}}$$ (4.26) Integrating $$y_0$$ using MATLAB's ode45 command, the following plot is obtained for y:
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Author
Solved and Typed By - Reviewed By -

Statement
Consider the following L2-ODE-CC; see p.6-6:
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$$\displaystyle{y}''-4{y}'+13{y}=2e^{-2x}\cos(3x)$$ (6.0) Homogeneous solution:
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$$\displaystyle{y}_{h}(x)=e^{-2x}[A\cos3x+B\sin3x]$$ (6.1) Particular solution:
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$$\displaystyle{y}_{p}(x)=xe^{-2x}[M\cos3x+N\sin3x]$$ (6.2) Complete the solution for this problem.
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Find the overall solution $$y(x)$$ that corresponds to the initial condition (3b) p.3-7
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$$\displaystyle y(0)=1,y'(0)=0$$ (6.3)
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Solution
Start by finding $$y_{p}'$$ and $$y_{p}''$$.
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$$\displaystyle {y}_{p}'={{e}^{-2x}}[(-3Mx-2Nx+N)\sin 3x+(-2Mx+3Nx+M)\cos 3x]$$ (6.4)
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$$\displaystyle {y}_{p}''={{e}^{-2x}}[(12Mx-6M-5Nx-4N)\sin 3x+(-15Mx-12M-12Nx+6N)\cos 3x]$$ (6.5) Substitute $$y_p$$ and its derivatives into (6.0) to find $$M$$ and $$N$$.
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$$\displaystyle {y}_{p}''+4{y}_{p}'+13{y}_{p}={{e}^{-2x}}[(-6M)\sin 3x+(-10Mx-8M+6N)\cos 3x]$$ (6.6) Separating terms and setting equal to the excitation from (6.0):
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$$\displaystyle -6M{{e}^{-2x}}\sin 3x+(-8M+6N){{e}^{-2x}}\cos 3x-10Mx{{e}^{-2x}}\cos 3x=2{{e}^{-2x}}\cos 3x$$
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(6.7) From (6.7), we solve coefficients to get
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$$\displaystyle -6M=0$$ $$\displaystyle -8M+6N=2$$ $$\displaystyle -10M=0$$ (6.8) Solving for $$M$$ and $$N$$:
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$$\displaystyle M=0$$ $$\displaystyle N=\frac{1}{3}$$ (6.9) Which gives us the particular solution:
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 * }
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$$\displaystyle {y}_{p}=\frac{1}{3}x{{e}^{-2x}}\sin 3x$$ (6.10) For the general solution,
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 * }
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$$\displaystyle y={{y}_{h}}+{{y}_{p}}$$ (6.11)
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 *

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$$\displaystyle y=e^{-2x}[A\cos3x+B\sin3x]+\frac{1}{3}x{{e}^{-2x}}\sin 3x$$ (6.12)
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 * }
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$$\displaystyle y=e^{-2x}[A\cos3x+(B+\frac{1}{3}x)\sin3x]$$ (6.13) To solve for $$A$$ and $$B$$, we use initial conditions from (6.3):
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$$\displaystyle y(0)=e^{-2(0)}[A\cos3(0)+(B+\frac{1}{3}(0))\sin3(0)]=1$$ (6.14) Which simplifies to:
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$$\displaystyle A=1$$ (6.15) For the second initial condition from (6.3):
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$$\displaystyle y'={{e}^{-2x}}[(-3A-2B+\frac{1}{3}+\frac{2}{3}x)\sin 3x+(3B+x-2)\cos 3x]$$ (6.16)
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$$\displaystyle y'(0)={{e}^{-2(0)}}[(-3A-2B+\frac{1}{3}+\frac{2}{3}(0))\sin 3(0)+(3B+(0)-2)\cos 3(0)]=0$$ (6.17)
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$$\displaystyle 3B-2=0$$ (6.18)
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$$\displaystyle B=\frac{2}{3}$$ (6.19) We can now write $$y$$ with all coefficients known.
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$$\displaystyle y={{e}^{-2x}}[(1)\cos 3x+(\frac{2}{3}+\frac{1}{3}x)\sin 3x]$$ (6.20) Final Equation
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$$\displaystyle y={{e}^{-2x}}[\cos 3x+(\frac{2}{3}+\frac{1}{3}x)\sin 3x]$$ (6.21)
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 *

 * }

Author
Solved and Typed By - Egm4313.s12.team1.essenwein (talk) 23:45, 19 March 2012 (UTC) Reviewed By -

title: User:Egm4313.s12.team1.essenwein


text:
User:Egm4313.s12.team18.hollon

=William Hollon=

Useful Links
Team 18 Course Wiki Course Website Report Table

Problem
The goal of this exercise is to derive the alternative formulations of the circuit equation seen in (3) and (4) using the circuit equation from (2). Where:

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$$  \displaystyle Q=Cv_C \Rightarrow \int i\,dt=Cv_c \Rightarrow i = C \frac{dv_C}{dt} $$     (1a,1b,1c)
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 *
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$$  \displaystyle V=LC\frac{d^2v_C}{dt^2}+RC\frac{dv_C}{dt}+v_C $$     (2)
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 *
 * }

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$$  \displaystyle LI''+RI'+\frac{1}{C}I=V' $$     (3)
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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4)
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 *
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Deriving (3) from (2)
In order to derive first alternate version, (3), of the circuit equation, (2), we begin by taking the derivative of (2) with respect to time, the result of which can be seen below in (5)

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$$  \displaystyle V'=LC \frac{d^3 v_C}{dt^3} + RC \frac{d^2 v_C}{dt^2} + \frac{d v_C}{dt} $$     (5)
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Next, by rearranging the right hand equation from (1) we obtain the following relation:

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$$  \displaystyle \frac{I}{C}=\frac{d v_C}{dt} $$     (6)
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And by taking subsequent derivatives, we also get:
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$$  \displaystyle \frac{I'}{C}=\frac{d^2 v_C}{dt^2} \, \, \& \, \, \frac{I''}{C}=\frac{d^3 v_C}{dt^3} $$     (7) & (8)
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By plugging the relations from (6), (7), and (8) into equation (2) and reducing the fractions we obtain the variant seen in (3)
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$$  \displaystyle LI'' + RI' + \frac{1}{C}I=V' $$     (3)
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Deriving (4) from (2)
We can derive the second alternate equation, (4), from circuit equation (2) using a method similar to the one used to derive (3) as shown above.

By taking the derivatives of (1a) we obtain the following relations:

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$$  \displaystyle Q=Cv_C \, \, \& \, \, Q'=C \frac{dv_C}{dt} \, \, \& \, \, Q''=C \frac{d^2 v_C}{dt^2} $$     (9a, 9b, 9c)
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Once again, plugging (9a, 9b, 9c) into (2) and reducing the results we can obtain the second variation of the circuit equation: (4)
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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4)
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 *
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title: User:Egm4313.s12.team18.hollon


text:
User:Egm4313.s12.team8.dupre

Egm4313.s12.team8.dupre/R1.2

Egm4313.s12.team8.dupre/R2.3

Egm4313.s12.team8.dupre/R2.7

Egm4313.s12.team8.dupre/R3.3

Egm4313.s12.team8.dupre/R3.4

Egm4313.s12.team8.dupre/R4.4.3

Egm4313.s12.team8.dupre/R5.4

Egm4313.s12.team8.dupre/R6.1

Egm4313.s12.team8.dupre/R7.3

title: User:Egm4313.s12.team8.dupre


text:
User:Egm4313.s12.team2/Report1

= R1.1 =

Problem Statement
Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$f(t)$$. Figure 1.1: The spring is represented by K, the dashpot is represented by C, the mass is represented by M, and the applied force is represented by $$\bar{r}(t)$$. --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Kinematics

 * $$\bar{r}=y\bar{E}_y$$
 * $$\bar{v}={y}'\bar{E}_y$$
 * $$\bar{a}=y''\bar{E}_y$$

Description of Kinematics
The kinematics describing the motion of the spring-dashpot-mass system in Fig. 1. are derived, first, from the position function of the mass with respect to time ($$r(t)$$). The next expression, $$\bar{v}={y}'\bar{E}_y$$,is the function for velocity, which is the derivative of the position function $$r(t)$$. The expression, $$\bar{a}=y''\bar{E}_y$$ represents the acceleration of the mass with respect to time, and is the derivative of the velocity function, $$\bar{v}={y}'\bar{E}_y$$.

Kinetics- Free Body Diagram
Figure 1.2 --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Using Newton's Second Law

 * $$F=m\bar{a}$$

Summing Forces
 * $$\bar{F}_k+\bar{F}_c+m\bar{g}=r(t)$$

$$ky\bar{E}_y+cy'\bar{E}_y+my''\bar{E}_y=r(t)$$ --Corey Williams 20:41, 27 January 2012 (UTC)

=R1.2=

Problem Statement
Derive the equation of motion of the spring-mass-dashpot in figure 53. on page 85 of Kreyszig's Advanced Engineering Mathematics (2011). Figure 2.1 --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Kinematics

 * $$ \bar{r} = y \bar{E}_y $$
 * $$ \bar{v}=y' \bar{E_y}$$
 * $$ \bar{a} = \bar{v}' \bar{E_y}=y''\bar{E_y}$$

Description of Kinematics
The kinematics describing the motion of the spring-dashpot-mass system in Fig. 53 (also pictured above) are derived, first, from the position function of the mass with respect to time $$\bar{r}(t) = y\bar{E}_y $$. The next expression, $$\bar{v}={y}'\bar{E}_y$$,is the function for velocity, which is the derivative of the position function $$\bar{r}(t) = y\bar{E}_y $$. The expression, $$\bar{a}=y''\bar{E}_y$$ represents the acceleration of the mass with respect to time, and is the derivative of the velocity function, $$\bar{v}={y}'\bar{E}_y$$.

Kinetics-Free Body Diagram
Figure 2.2 --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Using Newton's Second Law
Equations:
 * $$\bar{F} = m\bar{a} $$
 * $$\bar{F}_{spring} = ky\bar{E}_y$$
 * $$\bar{F}_{dashpot} = cy'\bar{E}_y $$
 * $$ mg = m\bar{a} = my''\bar{E}_y$$

Sum the Forces
 * $$ ky\bar{E}_y + cy'\bar{E}_y + my''\bar{E}_y = r(t) $$

From Kreyszig pg. 86
 * $$ \bar{r}(t) = \bar{F}_0{cos}(wt) $$

$$ ky\bar{E}_y + cy'\bar{E}_y + my''\bar{E}_y = \bar{F}_0{cos}(wt) $$

--Stephen Leshko 20:10, 30 January 2012 (UTC)

=R1.3=

Problem Statement
For the spring-dashpot-mass system on p.1-4, draw the FBDs and derive the equation of motion (2)p.1-4. Figure 3.1 --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Free Body Diagrams
Figure 3.2 FBD of spring, dashpot, and mass --Rachel Dzadek 22:20, 31 January 2012 (UTC) Figure 3.3 --Rachel Dzadek 22:20, 31 January 2012 (UTC)

Position Function

 * $$ y(t)=y_k + y_c                    \ $$

 * $$ y'(t)=y'_k + y'_c                 \ $$

 * $$ y(t)=y_k + y''_c              \ $$

From Figure 3.2

$$ f(w)=f(k)=f(c)=:\ f(I)              $$

From Figure 3.3

$$ \overline{f}(t)-\overline{f}(I)=my'' $$

$$ \overline{f}(t)=\overline{f}(I)+my'' $$

--Corey Williams 18:29, 1 February 2012 (UTC)

=R1.4=

Problem Statement
Derive (3) and (4) on p.2-2 from (2) on the same page.

 * $$ LC\frac{{d^2}v_c }{dt^2} + RC\frac{dv_c}{dt} + v_c = V \ $$ .............(2)

 * $$ LI'' + RI' + \frac{1}{C}I = V' \ $$ .............(3)

 * $$ LQ'' + RQ' + \frac{1}{C}Q = V \ $$ ...........(4)

Useful Equations
The two equations can be derived using:
 * $$ Q = Cv_c ; Q' = Cv'_c; Q= Cv_c \ $$ ......(5)
 * $$ I = \frac{dQ}{dt}=Q' ; I' = Q ; I = Q''' \ $$ ..... (6)

Derivations
Also knowing that equation (2) can be written as:
 * $$ V = LCv_c'' +RCv_c' + v_c \ $$ .....(2')

Then by the simple substitutions in (5) equation (2) becomes: $$ V = LQ'' + RQ' + \frac{1}{C}Q \ $$ ....... (4) Then by taking the derivative with respect to time, and again treating R, L, and C as constants equation (2) becomes:
 * $$ V' = LQ' + RQ +\frac{1}{C}Q' \ $$ ......(7)

Then by using the substitutions in (6) equation (7) becomes: $$ V' = LI'' +RI' + \frac{1}{C}I \ $$ .....(3)

--Stephen Leshko 22:32, 31 January 2012 (UTC)

=R1.5=

Problem Statement
Do problems 4-5 on page 59 of Kreyszig's Advanced Engineering Mathematics (2011).

Problem (4)

 * $$ y''+ 4y'+(\pi^2+4)y=0 \ $$

 * $$ r^2+4r=(\pi^2+4)=0 \ $$

 * $$ r=(\frac{-4\pm \sqrt{16-4(\pi^2 +4)}}{2}) \ $$

 * $$r=-2\pm sqrt{4-(\pi^2+4)} \ $$

 * $$r=-2\pm\pi i \ $$

Therefore, the general solution takes the form of:
 * $$y_g=Ae^{-2t}cos(\pi t)+Be^{-2t}sin(\pi t) \ $$

 * $$y'=A[-2e^{-2t}cos(\pi t)-\pi e{^-2t}sin(\pi t)]+B[-2e^{-2t}sin(\pi t)+\pi e^{-2t}cos(\pi t)] \ $$

 * $$y''=A[4e^{-2t}cos(\pi t)+2\pi e{^-2t}sin(\pi t)+2\pi e^{-2t}sin(\pi t)-\pi^2 e^{-2t}cos(\pi t)] \ $$

 * $$ +B[4e^{-2t}sin(\pi t)-2\pi e^{-2t}cos(\pi t)-2\pi e^{-2t}cos(\pi t)-\pi^2 e{^-2t}sin(\pi t)] \ $$

Substitution back into the original equation should result in 0=0
 * $$ 0 = A[4e^{-2t}cos(\pi t)+2\pi e{^-2t}sin(\pi t)+2\pi e^{-2t}sin(\pi t)-\pi^2 e^{-2t}cos(\pi t)] \ $$
 * $$+B[4e^{-2t}sin(\pi t)-2\pi e^{-2t}cos(\pi t)-2\pi e^{-2t}cos(\pi t)-\pi^2 e{^-2t}sin(\pi t)]  \ $$
 * $$+(4A[-2e^{-2t}cos(\pi t)-\pi e{^-2t}sin(\pi t)]+B[-2e^{-2t}sin(\pi t)+\pi e^{-2t}cos(\pi t)] +(\pi^2+4)(Ae^{-2t}cos(\pi t)+Be^{-2t}sin(\pi t))                                           \ $$
 * $$ 0=0                                                                                \ $$

Substitution of the solution and its derivatives into the original differential equation returns a value of zero, thereby proving that the solution is correct. --Nicolas Ellis 20:10, 30 January 2012 (UTC)

Problem (5)

 * $$ y'' +2\pi y' +\pi y = 0 \ $$

 * $$ r^2 +2\pi r +\pi ^2 = 0 \ $$

 * $$r = \frac{-2\pi \pm \sqrt{4\pi ^2 - 4\pi^2} }{2} \ $$

 * $$ r= \frac{-2\pi}{2} = -\pi\ $$

Therefore, the general solution takes the form of:
 * $$ y_g = C_1e^{-\pi t} + C_2te^{-\pi t} \ $$

 * $$ y' = -\pi C_1e^{-\pi t} + C_2[e^{-\pi t} -\pi te^{-\pi t}] \ $$

 * $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi e^{-\pi t} - \pi(e^{-\pi t}-\pi te^{-\pi t})] \ $$

 * $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi ^2te^{-\pi t} - 2\pi e^{-\pi t}] \ $$

Substitution back into the original equation should result as 0=0

 * $$ 0 = \pi^2C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} -2\pi C_2e^{-\pi t} - 2\pi^2C_1e^{-\pi t} + 2\pi C_2e^{-\pi t} -2\pi^2C_2e^{-\pi t}+\pi^2 C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} \ $$
 * $$ 0 = 0 \ $$

Thus the substitution checks out, and the form of the general solution is proven. --Stephen Leshko 20:10, 30 January 2012 (UTC)

=R1.6=

Problem Statement
For each ODE in Fig.2 in K 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

Falling Stone

 * $$ y'' = g = const. \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h '' = 0                       \ $$
 * $$ y_p '' = g                       \ $$
 * $$ y(t) := (y_h + y_p)          \ $$
 * $$ y_h  + y_p  = 0 + g  \ $$
 * $$ y(t)'' = g                       \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Parachute

 * $$ mv' = mg - bv^2 \ $$

Order: 1st order Linearity (or lack of): not linear Superposition:
 * $$ mv_h ' + kv_h ^2 = 0                 \ $$
 * $$ mv_p ' + kv_p ^2 = mg                \ $$
 * $$ y(x) := y_h + y_p                          \ $$

 * $$ y(x) = mv_h ' + kv_h ^2 + mv_p ' + kv_p ^2 \ $$
 * $$ y_h + y_p = mg                             \ $$

 * $$ mv_h ' + kv_h ^2 + mv_p ' + kv_p ^2 = mg   \ $$
 * $$ m(v_h ' + v_p ') + k(v_h ^2 + v_p ^2) = mg \ $$
 * $$ m(v_h + v_p)' + k(v_h + v_p)^2 \neq mg     \ $$

Superposition false $$ \therefore $$ not linear. --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Outflowing Water

 * $$ h' = -k \sqrt(h) \ $$

Order: 1st order Linearity (or lack of): not linear Superposition :
 * $$ y_h = h_h ' + k \sqrt h_h = 0         \ $$
 * $$ y_p = h_p ' = -k \sqrt h_p            \ $$
 * $$ y(x) := y_h + y_p                     \ $$

 * $$ y(x) = h_h ' + k \sqrt h_h + h_p '    \ $$
 * $$ y_h + y_p = -k \sqrt h_p              \ $$

 * $$ \frac {dh} {dt} = -kh^ \frac 1 2      \ $$
 * $$ \int h^ {- \frac 1 2} dh = - \int k dt \ $$
 * $$ 2h^ \frac 1 2 = -kt + c               \ $$
 * $$ h^ \frac 1 2 = - \frac {kt} 2 + c     \ $$
 * $$ h = ( - \frac {kt} 2 + c )^2          \ $$

 * $$ h(1) = ( - \frac k 2 + c )^2          \ $$
 * $$ h(2) = ( -k + c )^2                   \ $$
 * $$ h(3) = ( - \frac {3k} 2 + c )^2       \ $$
 * $$ h(1) \neq h(2) \neq h(3)              \ $$

Superposition false $$ \therefore $$ not linear. --Rachel Dzadek 23:00, 31 January 2012 (UTC)

Vibrating Mass on a Spring

 * $$ my'' + ky = 0 \ $$

Order: 2nd order Linearity (or lack of): linear Superposition :
 * $$ my'' + ky = 0                                                      \ $$
 * $$ y'' = - \frac k m y                                                \ $$

Let $$ y_1 \ $$ and $$ y_2 \ $$ be solutions.
 * $$ y_1 '' = - \frac k m y_1                                           \ $$
 * $$ y_2 '' = - \frac k m y_2                                           \ $$

 * $$ (ay_1 + by_2)'' = - \frac k m (ay_1 + by_2)                        \ $$
 * $$ ay_1  + by_2  = - \frac k m (ay_1 + by_2)                      \ $$
 * $$ a(- \frac k m y_1) + b(- \frac k m y_2) = - \frac k m (ay_1 + by_2) \ $$
 * $$ - \frac k m (ay_1 + by_2) = - \frac k m (ay_1 + by_2)              \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

Beats of a Vibrating System

 * $$ y'' + \omega _0 ^2 y = \cos \omega t \ $$
 * $$ \omega_0 = \omega \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h = y_h '' + \omega _0 ^2 y_h = 0                                  \ $$
 * $$ y_p = y_p '' + \omega _0 ^2 y_p = \cos \omega t                      \ $$
 * $$ y(x) := y_h + y_p                                                    \ $$

 * $$ y(x) = y_h  + \omega _0 ^2 y_h + y_p  + \omega _0 ^2 y_p         \ $$
 * $$ y_h + y_p = 0 + \cos \omega t                                        \ $$

 * $$ y_h  + \omega _0 ^2 y_h + y_p  + \omega _0 ^2 y_p = \cos \omega t \ $$
 * $$ (y_h  + y_p ) + \omega _0 ^2 (y_h + y_p) = \cos \omega t         \ $$
 * $$ (y_h + y_p)'' + \omega _0 ^2 (y_h + y_p) = \cos \omega t             \ $$
 * $$ y'' + \omega _0 ^2 y = \cos \omega t                                 \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

Current I in an RLC Circuit

 * $$ LI'' + RI' + \frac 1 C I = E' \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h = LI_h '' + RI_h ' + \frac 1 C I_h = 0                                        \ $$
 * $$ y_p = LI_p '' + RI_p ' + \frac 1 C I_p = E'                                       \ $$
 * $$ y(x) := y_h + y_p                                                                 \ $$

 * $$ y(x) = LI_h  + RI_h ' + \frac 1 C I_h + LI_p  + RI_p ' + \frac 1 C I_p        \ $$
 * $$ y_h + y_p = 0 + E'                                                                \ $$

 * $$ LI_h  + RI_h ' + \frac 1 C I_h + LI_p  + RI_p ' + \frac 1 C I_p = E'          \ $$
 * $$ L(I_h  + I_p ) + R(I_h ' + I_p ') + \frac 1 C (I_h + I_p) = E'                \ $$
 * $$ L(I_h + I_p)'' + R(I_h + I_p)' + \frac 1 C (I_h + I_p) = E'                       \ $$
 * $$ LI'' + RI' + \frac 1 C I = E'                                                     \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

Deformation of a Beam

 * $$ EIy^{IX} = f(x) \ $$

Order: 4th order Linearity (or lack of): linear Superposition:
 * $$ y_h = EIy_h ^{IX} = 0           \ $$
 * $$ y_p = EIy_p ^{IX} = f(x)        \ $$
 * $$ y(x) := y_h + y_p               \ $$

 * $$ y(x) = EIy_h ^{IX} + EIy_p ^{IX} \ $$
 * $$ y_h + y_p = 0 + f(x)            \ $$

 * $$ EIy_h ^{IX} + EIy_p ^{IX} = f(x) \ $$
 * $$ EI(y_h ^{IX} + y_p ^{IX}) = f(x) \ $$
 * $$ EI(y_h + y_p)^{IX} = f(x)       \ $$
 * $$ EIy^{IX} = f(x)                 \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

Pendulum

 * $$ L \theta '' + g \sin \theta = 0 \ $$

Order: 2nd Linearity (or lack of): not linear Superposition:
 * $$ y_h = L \theta _h '' + g \sin \theta_h = 0                                      \ $$
 * $$ y_p = L \theta _p '' + g \sin \theta_p = 0                                      \ $$
 * $$ y(x) := y_h + y_p                                                               \ $$

 * $$ y(x) = L \theta _h  + g \sin \theta_h + L \theta _p  + g \sin \theta_p      \ $$
 * $$ y_h + y_p = 0 + 0                                                               \ $$

 * $$ L \theta _h  + g \sin \theta_h + L \theta _p  + g \sin \theta_p = 0         \ $$
 * $$ L( \theta _h  + \theta _p ) + g( \sin \theta_h + \sin \theta_p) = 0         \ $$
 * $$ L( \theta _h + \theta _p)'' + g( \sin \theta_h + \sin \theta_p) = 0             \ $$
 * $$ L \theta  + g( \sin \theta_h + \sin \theta_p) \neq L \theta  + g \sin \theta \ $$

Superposition false $$ \therefore $$ not linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

=Effort Distribution=

--Corey Williams 15:58, 31 January 2012 (UTC)

title: User:Egm4313.s12.team2/Report1


text:
User:Egm4313.s12.team11.imponenti

Reports

title: User:Egm4313.s12.team11.imponenti


text:
User:Egm4313.s12.team2.dzadek.rl

About Me
My name is Rachel Dzadek. I am a mechanical engineering student at the University of Florida. I am in Team 2

R1.6
QUESTION: For each ODE in Fig.2 in K 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

Falling Stone

 * $$ y'' = g = const. \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h '' = 0                       \ $$
 * $$ y_p '' = g                       \ $$
 * $$ y(x) := (y_h + y_p)          \ $$

 * $$ y(x) = y_h  + y_p '' = 0 + g \ $$
 * $$ y(x) = (y_p + y_h)           \ $$
 * $$ y(x)'' = g                       \ $$

Superposition true $$ \therefore $$ linear.

Parachute

 * $$ mv' = mg - bv^2 \ $$

Order: 1st order Linearity (or lack of): not linear Superposition:
 * $$ y_h = mv_h ' + kv_h ^2 = 0                 \ $$
 * $$ y_p = mv_p ' + kv_p ^2 = mg                \ $$
 * $$ y(x) := y_h + y_p                          \ $$

 * $$ y(x) = mv_h ' + kv_h ^2 + mv_p ' + kv_p ^2 \ $$
 * $$ y_h + y_p = 0 + mg                         \ $$

 * $$ mv_h ' + kv_h ^2 + mv_p ' + kv_p ^2 = mg   \ $$
 * $$ m(v_h ' + v_p ') + k(v_h ^2 + v_p ^2) = mg \ $$
 * $$ m(v_h + v_p)' + k(v_h + v_p)^2 \neq mg     \ $$

Superposition false $$ \therefore $$ not linear.

Outflowing Water

 * $$ h' = -k \sqrt(h) \ $$

Order: 1st order Linearity (or lack of): not linear Superposition :
 * $$ y_h = h_h ' + k \sqrt h_h = 0         \ $$
 * $$ y_p = h_p ' = -k \sqrt h_p            \ $$
 * $$ y(x) := y_h + y_p                     \ $$

 * $$ y(x) = h_h ' + k \sqrt h_h + h_p '    \ $$
 * $$ y_h + y_p = 0 + -k \sqrt h_p          \ $$

 * $$ \frac {dh} {dt} = -kh^ \frac 1 2      \ $$
 * $$ \int h^ {- \frac 1 2} dh = - \int k dt \ $$
 * $$ 2h^ \frac 1 2 = -kt + c               \ $$
 * $$ h^ \frac 1 2 = - \frac {kt} 2 + c     \ $$
 * $$ h = ( - \frac {kt} 2 + c )^2          \ $$

 * $$ h(1) = ( - \frac k 2 + c )^2          \ $$
 * $$ h(2) = ( -k + c )^2                   \ $$
 * $$ h(3) = ( - \frac {3k} 2 + c )^2       \ $$
 * $$ h(1) \neq h(2) \neq h(3)              \ $$

Superposition false $$ \therefore $$ not linear.

Vibrating Mass on a Spring

 * $$ my'' + ky = 0 \ $$

Order: 2nd order Linearity (or lack of): linear Superposition :
 * $$ my'' + ky = 0                                                      \ $$
 * $$ y'' = - \frac k m y                                                \ $$

Let $$ y_1 \ $$ and $$ y_2 \ $$ be solutions.
 * $$ y_1 '' = - \frac k m y_1                                           \ $$
 * $$ y_2 '' = - \frac k m y_2                                           \ $$

 * $$ (ay_1 + by_2)'' = - \frac k m (ay_1 + by_2)                        \ $$
 * $$ ay_1  + by_2  = - \frac k m (ay_1 + by_2)                      \ $$
 * $$ a(- \frac k m y_1) + b(- \frac k m y_2) = - \frac k m (ay_1 + by_2) \ $$
 * $$ - \frac k m (ay_1 + by_2) = - \frac k m (ay_1 + by_2)              \ $$

Superposition true $$ \therefore $$ linear.

Beats of a Vibrating System

 * $$ y'' + \omega _0 ^2 y = \cos \omega t \ $$
 * $$ \omega_0 = \omega \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h = y_h '' + \omega _0 ^2 y_h = 0                                  \ $$
 * $$ y_p = y_p '' + \omega _0 ^2 y_p = \cos \omega t                      \ $$
 * $$ y(x) := y_h + y_p                                                    \ $$

 * $$ y(x) = y_h  + \omega _0 ^2 y_h + y_p  + \omega _0 ^2 y_p         \ $$
 * $$ y_h + y_p = 0 + \cos \omega t                                        \ $$

 * $$ y_h  + \omega _0 ^2 y_h + y_p  + \omega _0 ^2 y_p = \cos \omega t \ $$
 * $$ (y_h  + y_p ) + \omega _0 ^2 (y_h + y_p) = \cos \omega t         \ $$
 * $$ (y_h + y_p)'' + \omega _0 ^2 (y_h + y_p) = \cos \omega t             \ $$
 * $$ y'' + \omega _0 ^2 y = \cos \omega t                                 \ $$

Superposition true $$ \therefore $$ linear.

Current I in an RLC Circuit

 * $$ LI'' + RI' + \frac 1 C I = E' \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h = LI_h '' + RI_h ' + \frac 1 C I_h = 0                                        \ $$
 * $$ y_p = LI_p '' + RI_p ' + \frac 1 C I_p = E'                                       \ $$
 * $$ y(x) := y_h + y_p                                                                 \ $$

 * $$ y(x) = LI_h  + RI_h ' + \frac 1 C I_h + LI_p  + RI_p ' + \frac 1 C I_p        \ $$
 * $$ y_h + y_p = 0 + E'                                                                \ $$

 * $$ LI_h  + RI_h ' + \frac 1 C I_h + LI_p  + RI_p ' + \frac 1 C I_p = E'          \ $$
 * $$ L(I_h  + I_p ) + R(I_h ' + I_p ') + \frac 1 C (I_h + I_p) = E'                \ $$
 * $$ L(I_h + I_p)'' + R(I_h + I_p)' + \frac 1 C (I_h + I_p) = E'                       \ $$
 * $$ LI'' + RI' + \frac 1 C I = E'                                                     \ $$

Superposition true $$ \therefore $$ linear.

Deformation of a Beam

 * $$ EIy^{IX} = f(x) \ $$

Order: 4th order Linearity (or lack of): linear Superposition:
 * $$ y_h = EIy_h ^{IX} = 0           \ $$
 * $$ y_p = EIy_p ^{IX} = f(x)        \ $$
 * $$ y(x) := y_h + y_p               \ $$

 * $$ y(x) = EIy_h ^{IX} + EIy_p ^{IX} \ $$
 * $$ y_h + y_p = 0 + f(x)            \ $$

 * $$ EIy_h ^{IX} + EIy_p ^{IX} = f(x) \ $$
 * $$ EI(y_h ^{IX} + y_p ^{IX}) = f(x) \ $$
 * $$ EI(y_h + y_p)^{IX} = f(x)       \ $$
 * $$ EIy^{IX} = f(x)                 \ $$

Superposition true $$ \therefore $$ linear.

Pendulum

 * $$ L \theta '' + g \sin \theta = 0 \ $$

Order: 2nd Linearity (or lack of): not linear Superposition:
 * $$ y_h = L \theta _h '' + g \sin \theta_h = 0                                      \ $$
 * $$ y_p = L \theta _p '' + g \sin \theta_p = 0                                      \ $$
 * $$ y(x) := y_h + y_p                                                               \ $$

 * $$ y(x) = L \theta _h  + g \sin \theta_h + L \theta _p  + g \sin \theta_p      \ $$
 * $$ y_h + y_p = 0 + 0                                                               \ $$

 * $$ L \theta _h  + g \sin \theta_h + L \theta _p  + g \sin \theta_p = 0         \ $$
 * $$ L( \theta _h  + \theta _p ) + g( \sin \theta_h + \sin \theta_p) = 0         \ $$
 * $$ L( \theta _h + \theta _p)'' + g( \sin \theta_h + \sin \theta_p) = 0             \ $$
 * $$ L \theta  + g( \sin \theta_h + \sin \theta_p) \neq L \theta  + g \sin \theta \ $$

Superposition false $$ \therefore $$ not linear.

title: User:Egm4313.s12.team2.dzadek.rl


text:
University of Florida/Egm4313/s12.team8.dupre/R1.2

Problem Statement

Derive the equation of the spring-mass-dashpot in Fig.53 (shown above) in K 2011 p.85, with an applied force $$\displaystyle r(t) $$ on the ball.

Kinematics
$$\displaystyle y=y_{c}=y_{k} $$ (2-0)

Kinetics
$$\displaystyle r(t)=m y''+f_{c}+f_{k} $$ (2-1)

Where
$$\displaystyle f_{k}=k y_{k}$$ (2-2) and $$\displaystyle f_{c}=c y_{c}$$ (2-3) Plugging into (2-1) gives us: $$\displaystyle r(t)=my''+cy_{c}+ky_{k}$$ (2-4) Using (2-0) and substituting into (2-4), we find that the final equation of motion for the spring-mash-dashpot is: $$\displaystyle r(t)=my''+cy'+ky $$ (2-5)


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User:Egm4313.s12.team15.howe

hello

title: User:Egm4313.s12.team15.howe