Question

Starting from origin a body oscillates simple harmonically with a period of $4$s. After what time will its kinetic energy be $50\%$ of the total energy?

• A. $2$s
• B. $4$s
• C. $\frac{1}{2}$s
• D. $3$s

Answer 1

Answer: (C)

$\frac{1}{2}$s

The kinetic energy of the oscillating particle is given by
$k=\frac{1}{2}m{\omega }^2(A^2-y^2)=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$
At $y=0,K=\frac{1}{2}m{\omega }^2{A}^2=K_{max}$
If after time t, the KE becomes $50\%$ of $K_{max}$, then$\frac{50}{100}{K}_{max}=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$
$\Rightarrow \frac{1}{2}\left(\frac{1}{2}m{\omega }^2{A}^2\right)=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$
$\cos \omega t=\frac{1}{r^2}=\cos \frac{\pi }{4}$
$\Rightarrow \omega t=\frac{\pi }{4}\Rightarrow \frac{2\pi }{T}t=\frac{\pi }{4}$
$t=\frac{T}{8}=\frac{4}{8}=\frac{1}{2}s$.