Finding displacement from the equilibrium

1. Nov 6, 2009

tjohn101

1. The problem statement, all variables and given/known data
Give the following answer in terms of the amplitude, A. At what displacement from equilibrium is the energy of a Simple Harmonic Oscillator two-thirds KE and one-third PE?

2. Relevant equations
I would guess
PE=.5(k)(x)^2
KE=.5(m)(v)^2
E=.5(k)(x)^2 + .5(m)(v)^2
If one is 2/3, would I change the 1/2 in it's particular equation to 2/3? What would I do from there?

3. The attempt at a solution
So far I have no idea where to even begin.

2. Nov 6, 2009

dotman

Hello,

Let me see if I can lead you in the right direction. You've got a couple of formulas there for the kinetic energy and the potential energy-- these can be defined further for simple harmonic motion, since the functions x(t) and v(t) are known.

Once you have these formulas, which contain the amplitude, you can set the 2/3 / 1/3 ratio and find x.

Have a look at the wikipedia page-- specifically, scroll down to Dynamics of simple harmonic motion and then, most importantly, Energy of simple harmonic motion.

http://en.wikipedia.org/wiki/Simple_harmonic_motion" [Broken]

Post again if you're still stuck. Good luck!

Last edited by a moderator: May 4, 2017
3. Nov 6, 2009

tjohn101

Hello and thank you for the answer!

Unfortunately, I'm still a bit stuck. There is an example in the book that is the exact same question but the energy is split 50/50 instead of 1/3 and 2/3. They got an answer of +/-.707.

How they got to that answer is completely beyond me.

4. Nov 6, 2009

dotman

I'm sorry, this problem is actually easier than I thought it was.

EDIT TO ADD: I just want to add, that by the above, I don't mean to imply that the problem is easy or hard or whatever. Only that you don't have to go into a whole lot of detail with the sin/cos stuff that I originally thought you did to arrive at the answer.

You have given the following equations in your first post:

$$U = \frac{1}{2}kx^2$$
$$K = \frac{1}{2}mv^2$$

$$E = U + K = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$$

Now if only your total energy formula was expressed in a form that had the Amplitude, you would have an equation that you can use to determine x or v (since you know the ratio you want).

Fortunately, you can do just that, using the equations from the wiki page I mentioned. The important result, for you, is that:

$$E = U + K = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \frac{1}{2}kA^2$$

Last edited: Nov 6, 2009
5. Nov 6, 2009

tjohn101

Okay I understand that, but how do I turn the .5(k)(A)^2 into an answer?

6. Nov 6, 2009

dotman

I've given you the missing piece you need. What have you tried?

You have a formula for the potential energy. You have a formula for the total energy. You have a ratio of the total energy to the potential energy, for this particular problem.

How can you put it all together?

The reason I ask is that going any further will amount to basically giving you the answer.

7. Nov 6, 2009

tjohn101

Every time I do it I get 1/3*A... but that isn't right.

8. Nov 6, 2009

rl.bhat

Displacement in a SHM is given by
x = Asinωt
Velocity will be
v = dx/dt = Aωcos(ωt) = ωsqrt(A^2 - x^2) ans E = 1/2*m*(A^2 - x^2)---(1)
Velocity is maximum when x = 0. So vmax = Aω and Emax = 1/2*m*vmax^2 = 1/2*m*ω^2*A^2 ----(2)
Ratio of E/Eamx is given. From equation 1and 2 find x.

9. Nov 6, 2009

tjohn101

Okay this time I ended up with x=1.5A

Sound anywhere near correct?

10. Nov 7, 2009

rl.bhat

E = 2/3*Emax
E/Emax = 2/3
(A^2 - x^2)/A^2 = 2/3
Solve for x.

11. Nov 7, 2009

dotman

No. You should know this is incorrect physically. Why?

That said, rl.bhat's way is the complicated way to solve this. If you use the potential energy term, which is already in $x^2$, you don't need to calculate the velocity to get at x. For example, in the example you said where they want to know at what x the energy is split 50/50, you know 50% of the total energy is equal to the potential energy:

$$\frac{1}{2}(\frac{1}{2}kA^2) = \frac{1}{2}kx^2$$

$$\rightarrow \frac{1}{2}A^2 = x^2$$

$$\rightarrow \frac{1}{\sqrt2}A = x$$

$$\rightarrow \pm .707A \simeq x$$

You can use this method for any proportion of U and K.