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Question

(A) The potential energy $(U)$ of a particle in simple harmonic motion at displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force

Vedclass · Accepted Answer

$3: 1$

Vedclass · Answer

(A) The potential energy $(U)$ of a particle in simple harmonic motion at displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.Given that the displacement $x = \frac{A}{2}$,where $A$ is the amplitude,the potential energy is $U = \frac{1}{2} k (\frac{A}{2})^2 = \frac{1}{8} k A^2$.The total energy $(E)$ of the particle is $E = \frac{1}{2} k A^2$.The kinetic energy $(K)$ is given by $K = E - U = \frac{1}{2} k A^2 - \frac{1}{8} k A^2 = \frac{3}{8} k A^2$.The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\frac{3}{8} k A^2}{\frac{1}{8} k A^2} = \frac{3}{1}$.Thus,the ratio is $3: 1$.

Column-$I$	Column-$II$
$(a)$ $y = \frac{A}{\sqrt{2}}$	$(i)$ $K = \frac{3E}{4}$
$(b)$ $y = \frac{\sqrt{3}A}{2}$	$(ii)$ $K = \frac{E}{4}$
	$(iii)$ $K = \frac{E}{2}$

For a particle executing simple harmonic motion,the ratio of kinetic and potential energies at a point where displacement is one half of the amplitude is

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