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Question

(C) The amplitude of the particle performing $SHM$ is $A = 10 \ cm$.The instantaneous displacement of the particle is $x = 6 \ cm$.The kinetic energy

Vedclass · Accepted Answer

$16: 9$

Vedclass · Answer

(C) The amplitude of the particle performing $SHM$ is $A = 10 \ cm$.The instantaneous displacement of the particle is $x = 6 \ cm$.The kinetic energy $(K)$ of the particle is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.The potential energy $(U)$ of the particle is given by $U = \frac{1}{2} m \omega^2 x^2$.The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{A^2 - x^2}{x^2}$.Substituting the values,we get $\frac{K}{U} = \frac{10^2 - 6^2}{6^2} = \frac{100 - 36}{36} = \frac{64}{36}$.Simplifying the fraction,$\frac{K}{U} = \frac{16}{9}$.Thus,the ratio of kinetic energy to potential energy is $16: 9$.

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

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