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Q.

A particle describes $S H M$ in a straight line about $O$ .

Given image
If the time period of the motion is $T$ then its kinetic energy at $P$ be half of its peak value at $O$ , if the time taken by the particle to travel from $O$ to $P$ is

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a

$\frac{1}{4} T$

b

$\frac{1}{2} T$

c

$\frac{1}{8} T$

d

$\frac{1}{2 \sqrt{2}} T$

answer is D.

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Detailed Solution

$K E_{m a x} = \frac{1}{2} K A^{2} {(K E)}_{P} = \frac{1}{2} K E_{m a x} = \frac{1}{2} K (A^{2} - x^{2})$

$\Rightarrow x = \frac{A}{\sqrt{2}}$ or $θ = 45 °$

So, time $= \frac{T}{360 °} \times 45 ° = \frac{T}{8}$

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