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(D) The displacement of a particle starting from the mean position is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega = 2\pi/T$

Vedclass · Accepted Answer

$3: 1$

Vedclass · Answer

(D) The displacement of a particle starting from the mean position is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega = 2\pi/T$ is the angular frequency.At $t = T/6$,the displacement is $x = A \sin((2\pi/T) \cdot (T/6)) = A \sin(\pi/3) = A \sqrt{3}/2$.The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k (A \sqrt{3}/2)^2 = \frac{3}{8} k A^2$.The total energy $E$ is given by $E = \frac{1}{2} k A^2$.The kinetic energy $K$ is $K = E - U = \frac{1}{2} k A^2 - \frac{3}{8} k A^2 = \frac{1}{8} k A^2$.The ratio of potential energy to kinetic energy is $U/K = (3/8 k A^2) / (1/8 k A^2) = 3/1$.Thus,the ratio is $3: 1$.

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$

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