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(B) The total energy of a simple harmonic oscillator is $E = \frac{1}{2} k A^2$,where $A$ is the amplitude.At the extreme position,the phase $\phi = 0

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$\sqrt{\frac{3 m E}{2}}$

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(B) The total energy of a simple harmonic oscillator is $E = \frac{1}{2} k A^2$,where $A$ is the amplitude.At the extreme position,the phase $\phi = 0$ (or $\pi$). Let the displacement be $x = A \cos(\omega t)$.After a phase shift of $	heta = \frac{\pi}{3}$,the displacement becomes $x = A \cos(\frac{\pi}{3}) = \frac{A}{2}$.The potential energy at this position is $U = \frac{1}{2} k x^2 = \frac{1}{2} k (\frac{A}{2})^2 = \frac{1}{4} (\frac{1}{2} k A^2) = \frac{E}{4}$.The kinetic energy $K$ at this position is $K = E - U = E - \frac{E}{4} = \frac{3E}{4}$.Since kinetic energy $K = \frac{p^2}{2m}$,we have $\frac{p^2}{2m} = \frac{3E}{4}$.Solving for momentum $p$,we get $p^2 = \frac{6mE}{4} = \frac{3mE}{2}$.Therefore,$p = \sqrt{\frac{3mE}{2}}$.

$A$ simple pendulum of mass $m$ executes $S.H.M.$ with total energy $E$. If at an instant it is at one of the extreme positions,then its linear momentum after a phase shift of $\frac{\pi}{3} \, rad$ will be

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