Question

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

• A. 1/4 s
• B. 1/12 s
• C. 1/8 s
• D. 1/6 s

Hint:

SHM: $x = A\sin(\omega t)$. Kinetic energy: $\frac{1}{2}m\omega^2(A^2 - x^2)$. Total energy: $\frac{1}{2}m\omega^2 A^2$.

Answer 1

The Correct Option is C

Let the equation of SHM be $x = A\sin(\omega t)$, where $A$ is the amplitude and $\omega$ is the angular frequency. Since the particle starts from the mean position, the initial phase is 0. Kinetic energy $K = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 A^2 \cos^2(\omega t)$. Total energy $E = \frac{1}{2}m\omega^2 A^2$. The ratio of kinetic to total energy is $\frac{K}{E} = \cos^2(\omega t)$. We are given that $\frac{K}{E} = \frac{3}{4}$, so $\cos^2(\omega t) = \frac{3}{4}$. $\cos(\omega t) = \pm\frac{\sqrt{3}}{2}$. The time period $T = 1.5$ s, so $\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5} = \frac{4\pi}{3}$. $\omega t = \frac{\pi}{6}$, so $t = \frac{\pi/6}{4\pi/3} = \frac{1}{8}$ s. This is the minimum time.