Physics - SHM Question with Solution | TestHub

Given mass of the particle, $m=200\text{ g}=0.2\text{ kg}$

Amplitude, $A=0.2\text{ m}$

Kinetic energy at the mean position is the maximum kinetic energy, $K_{\text{max}}=16\times 10^{-3}\text{ J}$

The formula for maximum kinetic energy is:

$K_{\text{max}}=\frac{1}{2}m{\omega }^2{A}^2$

Substituting the given values:

$16\times 10^{-3}=\frac{1}{2}\times 0.2\times {\omega }^2\times (0.2{)}^2$

$16\times 10^{-3}=0.1\times {\omega }^2\times 0.04$

$16\times 10^{-3}=4\times 10^{-3}\times {\omega }^2$

${\omega }^2=\frac{16\times 10^{-3}}{4\times 10^{-3}}=4$

$\omega =2\text{ rad/s}$

The general equation for S.H.M. is $Y=A\sin (\omega t+\varphi )$.

Since the initial phase $\varphi =0^{\circ }$, the equation of motion becomes:

$Y=0.2\sin (2t)$

Answer: $Y=0.2\sin (2t)$