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A particle is executing SHM. At a point $x = \frac{A}{3}$ , kinetic energy of the particle is K, where A is the amplitude. At a point x = $\frac{2 A}{3}$ , kinetic energy of the particle will be:

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2 K

$K \sqrt{2}$

$\frac{5}{8} K$

$\frac{5}{3} K$

answer is C.

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Detailed Solution

$v = ω \sqrt{(A^{2} - x^{2})} \Rightarrow v \propto \sqrt{(A^{2} - x^{2})}$

At x = $\frac{A}{3}$

$\sqrt{(A^{2} - x^{2})} = \sqrt{(A^{2} - \frac{4 A^{2}}{9})} = A \sqrt{\frac{8}{9}}$

At x = $\frac{2 A}{3}$

$\sqrt{(A^{2} - x^{2})} = \sqrt{(A^{2} - \frac{4 A^{2}}{9})} = A \sqrt{\frac{5}{9}}$

$K^{'} = \frac{5}{9} \times \frac{9}{8} K = \frac{5}{8} K$

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