Question

A particle of mass $400\text{g}$ is executing SHM of amplitude $0.4\text{m}$. When it passes through the mean position, its kinetic energy is $32\times {10}^{-3}\text{J}$. If the initial phase of oscillation is $\frac{\pi }{4}$, then the equation of motion of the particle is

• A. $0.4\sin (2t+\frac{\pi }{4})$
• B. $0.4\cos (2t+\frac{\pi }{4})$
• C. $0.4\sin (t+\frac{\pi }{4})$
• D. $0.4\cos (t+\frac{\pi }{4})$

Answer 1

The correct option is C $0.4\sin (t+\frac{\pi }{4})$

Given,
Mass $=400\text{g}$
Amplitude $\left(A\right)=0.4\text{m}$
Kinetic energy at mean position $\left(KE\right)=32\times {10}^{-3}\text{J}$
Initial phase $\left(\varphi \right)=\frac{\pi }{4}$
As we know, $KE=\frac{1}{2}m{\omega }^2{A}^2$ (at mean position)
$32\times {10}^{-3}=\frac{1}{2}\times 0.4\times {\omega }^2\times (0.4{)}^2$
${\omega }^2=\frac{64\times {10}^{-3}}{0.4\times 0.4\times 0.4}$
$\Rightarrow \omega =1\text{rad/s}$
$\therefore$ Equation of motion can be written as
$x=A\sin (\omega t+\varphi )$
$=0.4\sin (t+\frac{\pi }{4})$