Are Kinetic and Potential Energies Equal in Simple Harmonic Motion?

[mindcircus]

Consider a simple harmonic oscillator. Calculate the time averages of the kinetic and potential energies over one cycle, and show that these quantities are equal. Next, calculate the space averages of the kinetic and potential energies.

I'm completely confused about these terms. Time average? Space average?

I do know that, with SHM, the equation of motion is -kx=ma, and w(omega)^2=k/m. The equation of motion becomes a+(w^2)x=0. The solution for this equation is x(t)=Asin(wt-delta). I used this x value for kinetic energy, (1/2)mv^2.

x(t)=Asin(wt-d)
v(t)=wAcos(wt-d)
T=(1/2)mv^2:
(1/2)m(w^2)(A^2)cos^2(wt-d)
Sub in w^2=k/m,
T=(1/2)k(A^2)cos^2(wt-d)

For potential energy,
U=(1/2)kx^2
U=(1/2)k(A^2)sin^2(wt-d)

Then, I found total energy, E=T+U.
E=(1/2)kA^2(cos^2(wt-d)+sin^2(wt-d))
E=(1/2)kA^2

The question cites that the average is over one cycle, which makes me think I should incorporate period, or frequency. Period=2pi*square root (m/k). I don't know where I'm going with this. I might have just done a bunch of superfluous calculations.

The answer for the time average is T=U=(m*A^2*w^2)/4. And the space average is U=(1/2)T=(m*A^2*w^2)/6. Actually, if someone could even explain what the question is asking for, it would be a great help.

Thank you so much!

 

[cookiemonster]

I imagine that the only difference between the time average and the space average would be the method used to get them. This means that we can either find the energy as a function of time and average it over a period or find the energy as a function of space and average it over the path taken in one cycle.

Also, note that the question does not ask for the total energy. It asks you to compare exclusively kinetic and exclusively potential average energies. In order to do that, you need to calculate the average value of each of these two. So here's a question: how would you find the average value of a random function? Was there a way you could do it that you learned in calculus class?

cookiemonster

 

[M98Ranger]

First of all, let's clarify the terms "time average" and "space average" in this context. In physics, the term "average" usually refers to the mean value of a quantity over a certain interval. In the case of a harmonic oscillator, the time average refers to the average value of a quantity over one complete cycle, while the space average refers to the average value over the entire space or distance of the oscillation.

Now, let's proceed with the calculations. As you correctly stated, the equation of motion for a simple harmonic oscillator is given by a + w^2x = 0, where a is the acceleration, w is the frequency (or angular frequency), and x is the displacement from equilibrium. The solution to this equation is x(t) = Asin(wt - d), where A is the amplitude and d is the phase angle.

To calculate the time averages of the kinetic and potential energies over one cycle, we need to integrate their expressions over one cycle. This can be done by using the period of the oscillator, T = 2π/w, as the limits of integration. So we have:

Time average of kinetic energy, T = (1/T)∫(1/2)mv^2 dt = (1/2)mA^2w^2(1/T)∫cos^2(wt - d) dt = (1/2)mA^2w^2(1/T)∫(1 + cos(2(wt - d))) dt

= (1/2)mA^2w^2(1/T)[t + (1/2w)sin(2(wt - d))] from t = 0 to t = T

= (1/2)mA^2w^2(T + (1/2w)sin(2wT - 2d) - (1/2w)sin(-2d)) = (1/2)mA^2w^2T

= (1/2)mA^2w^2(2π/w) = (1/2)kA^2

Similarly, the time average of potential energy can be calculated as:

Time average of potential energy, U = (1/T)∫(1/2)kx^2 dt = (1/2)kA^2(1/T)∫sin^2(wt - d) dt