# Simple harmonic motion

1. Feb 22, 2004

### denian

a particle of mass m moving in such manner that its displacement x varies with time t following equation x = a sin wt. ( w = omega )
where a and w are constants.
what is the average KE for the motion of the particle in one rev?

isnt the answer suppose to be 1/2 mw(sq)a(sq) ?

2. Feb 22, 2004

### mindcircus

In my physics class, to solve a problem like that, we used the equation KE=(1/2)m*v^2. So, when given the equation for x, take the derivative to get velocity, and insert it into the equation.

3. Feb 23, 2004

### denian

yeah, that is how they derive the equation
1/2 mw^2 a^2.

i use that way, but the answer is 1/4mw^2 a^2

4. Feb 23, 2004

### HallsofIvy

Staff Emeritus
The problem asked for the average kinetic energy.

Given a continuous function, its "average value" on an interval from a to b is $$\frac{\int_a^b f(x)dx}{b-a}$$

For this problem, the kinetic energy is (1/2)mv2= (1/2)ma2w2cos2(wt). One revolution will occur as wt goes from 0 to 2&pi; so as t goes from 0 to 2&pi;/w and so the integration is from 0 to 2&pi;/w.

The integral itself is ma2w2(&pi;/w) but then we need to divide by 2&pi/w-0: the average kinetic energy is (m/2)a2w.