1. ## Harmonic Motion

How to solve this:
6 a Calculate the displacement in terms of x0, expressed as a fraction of the amplitude x0, of a particle moving in simple harmonic motion with a speed equal to half the maximum value.
b Calculate the displacement at which the energy of the particle has equal amounts of kinetic and of potential energy
I am stuck at both (a) and (b) and just don't know how to do this. Thanks for your help in advance!.

2. ## Re: Simple Harmonic Motion

Originally Posted by Mark739
How to solve this:
6 a Calculate the displacement in terms of x0, expressed as a fraction of the amplitude x0, of a particle moving in simple harmonic motion with a speed equal to half the maximum value.
b Calculate the displacement at which the energy of the particle has equal amounts of kinetic and of potential energy
I am stuck at both (a) and (b) and just don't know how to do this. Thanks for your help in advance!.
assuming the object undergoing SHM starts at maximum displacement, $x_0$

$x(t) = x_0 \cos(\omega t)$, where $\omega = \dfrac{2\pi}{T}$

(a) $x'(t) = v(t) = -x_0 \cdot \omega \sin(\omega t) \implies |v_{max}| = x_0 \cdot \omega$

at half max speed, $\dfrac{x_0 \cdot \omega}{2} = |v(t)| = x_0 \cdot \omega \sin(\omega t) \implies \sin(\omega t) = \dfrac{1}{2} \implies \omega t = \dfrac{\pi}{6} \implies t = \dfrac{\pi}{6\omega}$

$x\left(\dfrac{\pi}{6\omega}\right) = x_0 \cos \left(\omega \cdot \dfrac{\pi}{6\omega}\right) = \dfrac{x_0 \sqrt{3}}{2}$

(b) Max kinetic energy (equal to maximum mechanical energy of the system) occurs at equilibrium and is $K_{max} = \dfrac{1}{2}m v_{max}^2 = \dfrac{1}{2}m (x_0 \omega)^2$

When kinetic and potential energy are equal, $K = \dfrac{1}{2}m|v(t)|^2 = \dfrac{1}{4}m (x_0 \omega)^2 \implies x_0 \omega \cos(\omega t) = \dfrac{x_0 \omega}{\sqrt{2}} \implies \omega t = \dfrac{\pi}{4} \implies t = \dfrac{\pi}{4\omega}$

$x\left(\dfrac{\pi}{4\omega} \right) = x_0 \cos\left(\omega \cdot \dfrac{\pi}{4\omega} \right) = \dfrac{x_0}{\sqrt{2}}$