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Energy in simple harmonic motion

  • Thread starter kjartan
  • Start date
15
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1. Homework Statement


This is my first post here.
I'm particularly unsure about b(ii)
Thanks for any and all replies!
I apologize in advance if I haven't used the correct conventions, but I hope that this is legible. I will learn the correct conventions for future posts but was pressed for time here.

Question:

A harmonic oscillator has angular frequency w and amplitude A.

(a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (assume that U = 0 at equilibrium)

(b) (i) How often does this occur in each cycle?
(ii) What is the time between occurrences?

(c) At an instant when the displacement is equal to A/2, what fraction of the total energy of the system is kinetic and what fraction is potential?


2. Homework Equations

E = (1/2)mv^2 + (1/2)kx^2 = (1/2)kA^2





3. The Attempt at a Solution


(a) (1/2)mv^2 = (1/2)kx^2 --> x = +/- A/rad(2)

v = rad(k/m)*x = rad(k/m)*[A/rad(2)]


(b) (i) 4 times

(ii) x = Acos(wt + phi), choose phi = pi/2

so, x = Asin(wt)
x/A= sin(wt)
t = (1/w)arcsin(x/A)

subst. from (a) gives us t = (1/w)arcsin(1/rad(2)) = (1/w)*(pi/4)

So, change in t = pi/(2w)

(c) E = K + U

U = (1/2)kx^2 = (!/2)k(A/2)^2 = k(A^2/8)

K = k(A^2/2) - k(A^2/8) = k(3A^2/8)

so, K/E = 3/4 --> U/E = 1/4
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,474
20
(a) (1/2)mv^2 = (1/2)kx^2 --> x = +/- A/rad(2)

v = rad(k/m)*x = rad(k/m)*[A/rad(2)]
OK. Note that you also have 2 solutions for v: [itex]v=\pm \sqrt{k/2m}A[/itex] (click on the equation if you want to see the code that generates it). Then you take the magnitude of each, as the problem asked you to. Since this is a 1D problem, the magnitude of a vector is just the absolute value.

(b) (i) 4 times
Agree.

(ii) x = Acos(wt + phi), choose phi = pi/2

so, x = Asin(wt)
x/A= sin(wt)
t = (1/w)arcsin(x/A)

subst. from (a) gives us t = (1/w)arcsin(1/rad(2)) = (1/w)*(pi/4)

So, change in t = pi/(2w)
I agree with your change in t, but I'm not so sure you want to use the arcsin function here. The arcsin function is too restrictive to capture all the solutions of the trigonometric equation, as its range is only [itex][-\pi /2,\pi /2][/itex].

By the way, you don't need to choose a value of [itex]\phi[/itex]. You could also do it like this.

[tex]A\cos(\omega t+\phi )=\pm\frac{A}{\sqrt{2}}[/tex]

[tex]\omega t+\phi=...\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}...[/tex]

[tex]\omega t+\phi=\frac{(2n+1)\pi}{4}[/tex], [tex]n\in\mathbb{Z}[/tex]

Now take the [itex]t[/itex] values for any two adjacent [itex]n[/itex], and subtract the smaller from the greater. [itex]\phi[/itex] will subract off without you having to choose a value for it.

(c) E = K + U

U = (1/2)kx^2 = (!/2)k(A/2)^2 = k(A^2/8)

K = k(A^2/2) - k(A^2/8) = k(3A^2/8)

so, K/E = 3/4 --> U/E = 1/4
That is correct.
 
15
0
Thanks much, Tom, your reply was most helpful and I appreciate it!

Thanks for the tip about LaTex, too.
 

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