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A particle undergoing simple harmonic motion has an amplitude of 10cm. When the particle is at rest a displacement of 6 cm from the center, then the ratio of its kinetic energy to potential energy is

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3 : 2

9 : 4

16 : 9

4 : 3

answer is C.

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Detailed Solution

$\begin{array}{l} A = 10 cm; y = 6 cm \\ \frac{K}{U} = \frac{\frac{1}{2} m ω^{2} (A^{2} - y^{2})}{\frac{1}{2} m ω^{2} y^{2}} \end{array}$

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