Hey guys.
Could someone demonstrate to me how to do this:
Show that for any particle moving in simple harmonic motion, the ratio of the average over one oscillation to the max. speed is 2: pi.
Do you mean the ratio of the average speed to the maximum speed?
No, you can't possibly mean that since an object in harmonic motion returns to its starting point after one oscillation and so its "average speed" is 0. What average is it that you are talking about?
Harmonic motion can be modeled as $\displaystyle x= A sin(\omega t)$. Its speed, then, is $\displaystyle v= x'= A\omega cos(\omega t)$. The maximum speed occurs when $\displaystyle cos(\omega t)$ takes its maximum value which is 1- the maximum speed is $\displaystyle A\omega$.
The average of a continuous function over an interval is its integral over that interval divided by the length of the interval. Here, an anti-derivative of v= x' is x itself so we just need to determine the interval. sin(T) has period $\displaystyle 2\pi$ so $\displaystyle sin(\omega t)$ completes one cycle when $\displaystyle \omega t= 2\pi$ or $\displaystyle t= \frac{2\pi}{\omega}$.
But, the integral of v= x' is $\displaystyle x= Asin(\omega t)$ which is 0 at both ends so, like I said before, its average is 0.
Actually, the average velocity is zero, not the average speed. The instantaneous speed of an object is the modulus of it's instantaneous velocity.
In this case, it's speed would be $\displaystyle |v| = A\omega\cos{\omega t}$
Integrating this from $\displaystyle t = 0 \ to \ \frac{2\pi}{\omega}$ , we get 4A. Dividing by the interval gives the average speed, that is $\displaystyle \frac{2A}{\pi} $.
Since the maximum speed is $\displaystyle A\omega $, the ratio of the average speed to the maximum speed over one oscillation becomes $\displaystyle \frac{2}{\pi \omega} $.
Either I've done something wrong, or the problem is wrong. This is equal to $\displaystyle \frac{2}{\pi} $ only if the frequency is one oscillation per second.