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Q.

A particle execute in S.H.M whose displacement equation is Y = 5 sin (10t–π /3). The minimum time after which the K.E of the particle will be maximum is

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a

$\frac{π}{30} \sec$

b

$\frac{π}{60} \sec$

c

$\frac{π}{90} \sec$

d

$\frac{π}{120} \sec$

answer is A.

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Detailed Solution

When y = 0;
velocity and KE are maximum
$\begin{array}{l} y = 5 \sin (10 t - π / 3) = 0 \\ \Rightarrow 10 t - π / 3 = 0; t = π / 30 \end{array}$

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A particle execute in S.H.M whose displacement equation is Y = 5 sin (10t–π /3). The minimum time after which the K.E of the particle will be maximum is