# Thread: Simple Harmonic Motion problem

1. ## Simple Harmonic Motion problem

Yet another unsolved homework problem, this one involving simple harmonic motion:

"The speed of a point moving along the x-axis is given by

v^2 = 108 + 72x - 36x^2

Find the mean position, the amplitude and the frequency of the motion."

What do they mean find the mean position? Shouldn't it be at the centre? I know that I can use the formula

x = Acos(ωt + α)

to find amplitude (A), and

f = 1/T where T = 2π/ω

to find frequency (f)

But I can't even seem to get started. Thanks for any help

2. at max displacement, $\displaystyle v = 0$

$\displaystyle 108 + 72x - 36x^2 = 0$

$\displaystyle 36(3 + 2x - x^2) = 0$

$\displaystyle 36(3-x)(1+x) = 0$

the particle oscillates between $\displaystyle x = 3$ and $\displaystyle x = -1$

because of symmetry, equilibrium will midway between these two x-values, at $\displaystyle x = 1$.

at equilibrium, the particle is at $\displaystyle v_{max} = A\omega$

$\displaystyle v^2 = 108 + 72(1) - 36(1^2) = 144$

$\displaystyle v = 12$

$\displaystyle 12 = A\omega = 2\omega$

$\displaystyle \omega = 6$

$\displaystyle f = \frac{\omega}{2\pi} = \frac{3}{\pi}$ hz

3. Cool thanks skeeter