Question

A point particle of mass $0.1kg$ executing SHM with amplitude of $0.1m$. When the particle passes through the mean position is KE is $8\times {10}^{-3}J$. Obtain the equation of motion of this particle if the initial phase of oscillation is ${45}^o$

Answer 1

Equation of motion of a particle in a simple harmonic motion is given by

$x=A\cos (\omega t+\varphi )$

Given

Amplitude, $A=0.1m$

Initial phase, $\varphi =45°$

$=\frac{\pi }{4}rad$

Mass of particle, $m=0.1kg$

Kinetic energy at the mean position, $E_k=8\times {10}^{-3}J$

So,

$\frac{1}{2}m{\omega }^2{A}^2=8\times {10}^{-3}$

${\omega }^2=8\times {10}^{-3}\left(\frac{2}{m{A}^2}\right)$

${\omega }^2=8\times {10}^{-3}\left(\frac{2}{0.1\times {0.1}^2}\right)$

$\omega =4s^{-1}$

The equation of motion is,

$s=0.1\cos (4t+\frac{\pi }{4})$