Question

The displacement of a particle executing SHM is half its amplitude. The fraction of its kinetic energy will be:

• A. 2/3
• B. 3/4
• C. 1/3
• D. 1/2

Answer 1

The correct option is B 3/4

x=A/2, thus $1/2=sin\omega t$ or $\omega t={30}^0$
Thus, $v=A\omega cos{30}^0=A\omega \sqrt{3}/2$
thus, $KE=0.5m{v}^2=0.5m.3/4\left(A^2{\omega }^2\right)=3/4\left(0.5m{A}^2{\omega }^2\right)=3/4K{E}_{max}$