Question

Starting from the origin, a body oscillates simple harmonically with a period of $2s$. After what time will its kinetic energy be $75\%$ of the total energy?

• A. $\frac{1}{12}s$
• B. $\frac{1}{6}s$
• C. $\frac{1}{4}s$
• D. $\frac{1}{3}s$

Answer 1

Answer: (B)

$\frac{1}{6}s$

Kinetic energy for SHM is $K=\frac{1}{2}m{\omega }^2(A^2-x^2)$ and total energy , $T=\frac{1}{2}m{\omega }^2{A}^2$

According to question, $\frac{1}{2}m{\omega }^2(A^2-x^2)=\left(75/100\right)\times \frac{1}{2}m{\omega }^2{A}^2$

or $A^2-\frac{3}{4}{A}^2=x^2$

or $x=A/2$

For a SHM, $x=A\sin \omega t$

or $A/2=A\sin \omega t$

or $\sin \pi /6=\sin \omega t$

or $\omega t=\pi /6$

or $\left(2\pi /T\right)t=\pi /6$

or $t=1/6s$ where time period given $T=2s$