Q.16. A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is 50% of total energy?
(cos45∘= \frac{1}{\sqrt{2}})
A. 0.1 s
B. 0.2 s
C. 0.4 s
D. 0.5 s
Answer :- D. 0.5 s
Explanation :-
T \cdot E=\frac{1}{2} \mathrm{kA}^2 \text{Kinetic energy is given by } \mathrm{K} \cdot \mathrm{E}=\frac{1}{2} \mathrm{k}\left(\mathrm{A}^2-\mathrm{x}^2\right) \text{K.E} = \frac{1}{2} \text{P.E} \ldots \text{(given)} \therefore \frac{1}{2} k\left(A^2-x^2\right)=\frac{1}{2}\left(\frac{1}{2} k A^2\right) \therefore \left(A^2-x^2\right)=\frac{1}{2} A^2 \therefore x=\frac{A}{\sqrt{2}} \text{The equation of displacement in SHM is} \mathrm{x}=\mathrm{A} \sin \frac{2 \pi \mathrm{t}}{\mathrm{~T}} \therefore \frac{\mathrm{~A}}{\sqrt{2}}=\mathrm{A} \sin \frac{2 \pi \mathrm{t}}{4} \ldots(\because \mathrm{~T}=4 \mathrm{sec}) \therefore \frac{1}{\sqrt{2}}=\sin \frac{\pi \mathrm{t}}{2} \therefore \frac{\pi \mathrm{t}}{2}=\sin ^{-1} \frac{1}{\sqrt{2}} \frac{\pi \mathrm{t}}{2}=\frac{\pi}{4} \therefore \mathrm{t}=0.5 \mathrm{~s}