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(B) The kinetic energy $(KE)$ of a particle in simple harmonic motion is given by $KE = \frac{1}{2} k A^2 \cos^2(\omega t)$,where $A$ is the amplitude

Vedclass · Accepted Answer

$\frac{T}{12}$

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(B) The kinetic energy $(KE)$ of a particle in simple harmonic motion is given by $KE = \frac{1}{2} k A^2 \cos^2(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.Given that the body starts from the mean position,the displacement is $x = A \sin(\omega t)$,and the velocity is $v = A \omega \cos(\omega t)$.The total energy $(TE)$ is $\frac{1}{2} k A^2$.We are given $KE = 75 \% 	ext{ of } TE = \frac{3}{4} TE$.So,$\frac{1}{2} k (A \cos(\omega t))^2 = \frac{3}{4} (\frac{1}{2} k A^2)$.$\cos^2(\omega t) = \frac{3}{4}$.$\cos(\omega t) = \frac{\sqrt{3}}{2}$.Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $\omega t = 30^{\circ} = \frac{\pi}{6} 	ext{ radians}$.Substituting $\omega = \frac{2\pi}{T}$,we get $\frac{2\pi}{T} t = \frac{\pi}{6}$.Solving for $t$,we get $t = \frac{T}{12}$.

Starting from the mean position,a body oscillates simple harmonically with a period $T$. After what time will its kinetic energy be $75 \%$ of the total energy? $(\sin 30^{\circ} = 0.5)$

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