(III) By direct substitution, show that Eq. 14–22, with Eqs. 1...

Question

(III) By direct substitution, show that Eq. 14–22, with Eqs. 14–23 and 14–24, is a solution of the equation of motion (Eq. 14–21) for the forced oscillator. [Hint: To find sin θ and cos θ from tan θ, draw a right triangle.]

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A pendulum of length L and mass M is subjected to a dampening force proportional to the velocity given by B multiplied by D theta PDT and A periodic driving torque given by torque of T is equal to Tau zero multiplied by cosine of Omega multiplied by T the equation of motion for the angular displacement theta is M multiplied by L squared multiplied by D squared theta by DT squared plus B multiplied by D theta by DT plus M multiplied by G multiplied by L multiplied by theta is all equal to tow zero, multiplied by cosine of omega. T determine what the solution to this equation will be given that A equals Tau or torque zero. So Tau zero divided by M multiplied by L squared multiplied by the square root of Omega squared minus G divided by L all to the power of two plus B squared multiplied by omega squared divided by M squared multiplied by L squared and Phi equals inverse tangent of G divided by L minus omega squared divided by omega multiplied by B divided by M multiplied by L squared. Awesome. So that's our end goal is we're trying to figure out what the solution to this equation is given the values for a, the amplitude and some angle phi Awesome. So we're also given some multiple choice answers for what the solution of the equation will be. And it's given to us as theta of T is equal to some equation. So let's read them off to see what our final answer might be. So A is Tau zero divided by M multiplied by L squared multiplied by omega squared minus G divided by L all multiplied by cosine of omega TB is Tau zero divided by M multiplied by L squared multiplied by the square root of Omega squared minus G divided by L all to the power two plus B multiplied by Omega divided by M multiplied by L squared all to the power two, all multiplied by the cosine of omega multiplied by T minus the inverse tangent of G divided by L minus omega squared divided by Omega multiplied by B divided by M multiplied by L square C is Tau zero divided by the square root of M multiplied by L squared multiplied by Omega squared plus G divided by L all to the power two, all multiplied by cosine of omega multiplied by T And finally, D is tau zero divided by M multiplied by G multiplied by L all multiplied by cosine of Omega multiplied by T minus inverse tangent of G divided by L minus omega squared divided by Omega. OK. So first off, let us note that we are given the equation of motion as, and I will box this in blue, M multiplied by L squared multiplied by D squared theta divi divided by DT or D squared. Theta by DT squared plus B multiplied by D theta by DT plus M multiplied by G multiplied by L multiplied by theta is all equal to Tau zero multiplied by the cosine of Omega T or Omega multiplied by T. So we can write this as so we can write M multiplied by L squared multiplied by D squared data by D E squared plus B multiplied by D theta by DT plus M multiplied by G multiplied by L multiplied by theta is all equal to Tau zero multiplied by cosine of Omega multiplied by T. So which we can rewrite this equation as D squared beta by DT square plus B divided by M multiplied by L squared. So elsewhere multiplied by D theta by DT plus G divided by L multiplied by theta is all equal to Tau zero divided by M multiplied by L squared multiplied by cosine of Omega multiplied by T or we can rewrite this equation as D squared theta by DT squared plus alpha multiplied by D beta by DT plus beta multiplied by beta is equal to F zero, multiplied by cosine of. So cos sign of omega multiplied by T. So let us note that in this case, and let's write this in blue and then let's do a little star here. This is a side note that alpha, in this case equals B divided by M multiplied by L squared and beta is equal to G divided by L. And let us also note that F zero is equal to Tau zero divided by M multiplied by L squared. OK. So moving right along here, let us note that the assumed solution for this particular equation of motion is so the assumed, so this is the assumed solution is theta T. So theta T is equal to a multiplied by cosine of omega multiplied by T minus Phi. OK. So, and we are given in the problem itself for the values of A and P. Thus, if we plug these values into our assumed motion equation, we will get, so we need to take our assumed solution and we need to plug in our, our values that are provided to us in the problem itself for A and five. So when we do that, we will get that theta of tea is equal to how zero divided by N multiplied by L squared multiplied by the square root of omega squared minus G divided by L. And this whole value is squared plus B multiplied by omega divided by and multiplied by L squared. And this values all to the power of two. Awesome. And then we need to multiply this expression by cosine of omega multiplied by T minus inverse tangent of. And don't forget your parentheses, inverse tangent of G divided by L minus omega squared, all divided by omega multiplied by be divided by M multiplied by L. And don't forget that L is squared. Awesome. So we got a little bit cramped here. So let me rewrite that. So nice. Like, so we can actually see it. So B divided by M multiplied by L squared. And don't forget your sets of parentheses, make sure we always have all of our parentheses together. And this is our final answer. Hooray, we did it. This problem is very easy to solve as long as you understand the, the physics that's associated with it. So if this seems unfamiliar to you, I would just say take a refresher course on the chapter in your textbook to read up on it. So you know the flow of how everything goes and that's it. So looking at our multiple choice answers, the correct answer has to be the letter B which states that theta of T is equal to Tau zero, divided by M multiplied by L squared multiplied by the square root of Omega squared minus G divided by L all the power of two plus B multiplied by Omega divided by M multiplied by L squared all to the power two, all multiplied by cosine of omega multiplied by T minus inverse tangent of G divided by L minus omega squared divided by Omega multiplied by B divided by M multiplied by L squared. And that's it. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

(III) By direct substitution, show that Eq. 14–22, with Eqs. 14–23 and 14–24, is a solution of the equation of motion (Eq. 14–21) for the forced oscillator. [Hint: To find sin θ and cos θ from tan θ, draw a right triangle.]

Key Concepts

Forced Oscillator

Equations of Motion

Trigonometric Functions

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