Waves, solving Energy with position?

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Homework Help Overview

The discussion revolves around a second-year physics problem involving wave motion and energy equations. The original poster is trying to understand how the equation for position, x=A cos(ωt+φ), relates to the energy equation E=1/2 m (dx/dt)² + 1/2 k x².

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the validity of the question's wording and whether it contains errors. There are suggestions to substitute the position equation into the energy equation to verify its correctness. Some participants express confusion about the relationship between the two equations and question the intent behind the problem.

Discussion Status

The conversation is ongoing, with participants offering various suggestions and expressing uncertainty about the problem's requirements. Some guidance has been provided regarding substituting the position equation into the energy equation, but there is no clear consensus on the best approach or understanding of the question.

Contextual Notes

There is mention of a potential typo in the energy equation, and participants are grappling with the implications of this on their understanding. The original poster also notes a feeling that the question may be incorrectly framed, which adds to the complexity of the discussion.

sjmacewan
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Howdy,

This is only for a second year course, but i didn't think it would get answered in the intro forums...

Ok, the wording of this question doesn't make much sense to me, but I assure you that this is what it says...
********************************************
Demonstrate how

[tex]x=A \cos (\omega t+\phi)[/tex]

can be the answer for

[tex]E=\frac{1}{2}m\frac{dx}{dt}^2 + \frac{1}{2}kx^2[/tex]

********************************************

I just don't see how this could possibly work. Any ideas would be helpful

-----------

As for my thoughts, I figured that either the question was written wrong, or that using the simplified energy eq'n (E=kA^2) would help...but I get nowheres using that still.
 
Last edited:
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sjmacewan said:
Demonstrate how

x=Acos(wt+phi)

can be the answer for

E=1/2m(d^2x/dt^2)^2 + 1/2k(x^2)
There's a typo in that last equation. d^2x/dt^2 is the acceleration, not the speed.
 
thanks for pointing that out, it was supposed to be velocity so i edited it to say such. I was mixing two questions together for a second...i still don't know what to do though
 
Just plug your equation for x into the energy equation and see if it makes sense as a solution.
 
no, that would just complicate things. when i actually subsititute the derivative in the first eq'n i can simplify it easily to [tex]E= \frac{1}{2} k A^2[/tex] but that doesn't answer the question...what it seems to want is for you to show how [tex]x=A \cos (\omega t+\phi)[/tex] actually IS the answer for [tex]E=\frac{1}{2}m\frac{dx}{dt}^2 + \frac{1}{2}kx^2[/tex]

which just sounds absurd to me.
 
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hmmmm 40+ views and nothing. I personally think that the question must be wrong, but I'll continue to check back in case anyone sees a relationship that I'm missing between the two expressions
 
Solve it as a differential equation? You have [tex]\frac{dx}{dt}^2[/tex] and [tex]x^2[/tex], so if you try [tex]x=A'e^{\lambda t}[/tex] it should work. Just a suggestion, I'm not 100% on this.
 
I'm a bit puzzled why you ignored my advice to just plug it in and see if it satisfies the equation. That's the simplest way to verify a proposed solution to any equation! And that seems to be all you are asked to do: Just demonstrate that the given function of x is a solution to the energy equation. And it is!
 
I'm sorry, I guess I just misunderstood what the question was asking really...
if you just plug it in and work though it you get the [tex]e=\frac{1}{2}kA^2[/tex] equation that one would expect, I just didn't think that THAT was what the question wanted.

Either way, sorry for ignorning your advise, but what you said is the only thing that I myself though about doing, and I wasn't convinced that was right.
 

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