The amplitude of a particle executing SHM about $O$ is $10 c m$ . Then

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Its total energy at any point is equal to its maximum K.E.

When the K.E. is $0.64$ of its max. K.E. its displacement is $6 c m$ from $O$ .

Its velocity is half the maximum velocity when its displacement is half the maximum displacement.

When the displacement is $5 c m$ from $O$ its K.E. is $0.75$ of its max. P.E.

answer is A, B, C.

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Detailed Solution

$E = K_{0} = U_{0} = \frac{1}{2} k A^{2} = \frac{1}{2} k {(0.1)}^{2}$

At $x = 0.06, U = \frac{1}{2} k {(0.06)}^{2}$

$K = \frac{1}{2} k {(0.1)}^{2} - \frac{1}{2} k {(0.06)}^{2} = 0.64 E$

$x = 0.05, U = \frac{1}{2} k {(0.05)}^{2}$

$K = \frac{1}{2} k {(0.1)}^{2} - \frac{1}{2} k {(0.05)}^{2} = 0.75 E$

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