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Question

(D) The total energy $E$ of a simple harmonic oscillator is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) The total energy $E$ of a simple harmonic oscillator is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.The potential energy $U$ at a displacement $y = \frac{A}{3}$ is given by $U = \frac{1}{2} k y^2 = \frac{1}{2} k (\frac{A}{3})^2 = \frac{1}{2} k \frac{A^2}{9} = \frac{E}{9}$.The kinetic energy $KE$ is the difference between total energy and potential energy: $KE = E - U = E - \frac{E}{9} = \frac{8E}{9}$.The ratio of total energy to kinetic energy is $\frac{E}{KE} = \frac{E}{\frac{8E}{9}} = \frac{9}{8}$.Comparing this to the given ratio $\frac{x}{8}$,we find $x = 9$.

When the displacement of a simple harmonic oscillator is one third of its amplitude,the ratio of total energy to the kinetic energy is $\frac{x}{8}$,where $x=$ . . . . . . .

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