Question

An object of mass 0.2 kg executes SHM along the x-axis with frequency $\left(25/\pi \right)Hz$.At the point x=0.04m the object has KE 0.5 J and PE 0.4.J.The amplitude of oscillation (in cm) is

Answer 1

$\omega =2\pi f=50$
m=0.2 kg
$x=Asin\omega t$
Thus, at x=0.04m, $sin\omega t=\frac{0.04}{A}$
$v=A\omega cos\omega t$
$KE=0.5m{v}^2=0.5J$ or $0.5=0.5\left(0.2\right)\left(A^2{\omega }^{2}co{s}^2\omega t\right)$
thus, $5=A^2\left(50{)}^2\right(1-{0.04}^2/A^2)$
Solve for A to get A=6 cm