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(C) The displacement of a particle in simple harmonic motion starting from the mean position is given by $x = A \sin(\omega t)$.The velocity of the pa

Vedclass · Accepted Answer

$\frac{1}{8} \ s$

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(C) The displacement of a particle in simple harmonic motion starting from the mean position is given by $x = A \sin(\omega t)$.The velocity of the particle is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.The kinetic energy $(K)$ is $K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t)$.The total energy $(E)$ is $E = \frac{1}{2}m A^2 \omega^2$.The ratio of kinetic energy to total energy is $\frac{K}{E} = \cos^2(\omega t)$.Given $\frac{K}{E} = \frac{3}{4}$,so $\cos^2(\omega t) = \frac{3}{4}$,which implies $\cos(\omega t) = \frac{\sqrt{3}}{2}$.This means $\omega t = \frac{\pi}{6}$.Given the time period $T = 1.5 \ s$,we have $\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5} = \frac{4\pi}{3} \ rad/s$.Substituting $\omega$ into the equation: $(\frac{4\pi}{3}) t = \frac{\pi}{6}$.Solving for $t$: $t = \frac{\pi}{6} 	imes \frac{3}{4\pi} = \frac{3}{24} = \frac{1}{8} \ s$.

$A$ particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is $1.5 \ s$,then the minimum time at which the ratio of the kinetic and total energies of the particle becomes $3: 4$ is

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