A particle starts oscillating simple harmonically form its mean position with time period $T$. At time…
A particle starts oscillating simple harmonically form its mean position with time period $T$. At time $t=\frac{T}{12}$, the ratio of the potential energy to kinetic energy of the particle is $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$
$1: 3$
$2: 1$
$3: 1$
$1: 2$
Solution
Let, the equation of the particle performing SHM is given by $x=A \sin \omega t$, where $\omega=\frac{2 \pi}{T}$
At $t=\frac{T}{12}, x=A \sin \left(\frac{2 \pi}{T} \cdot \frac{T}{12}\right)=A \sin \left(\frac{\pi}{6}\right)=\frac{A}{2}$
The potential energy of the particle at $T=\frac{T}{12}$
The kinetic energy of the particle at $t=\frac{T}{12}$
$\begin{aligned} & K=\frac{1}{2} m \omega^2 x^2\left(A^2-x^2\right)=\frac{1}{2} m \omega^2\left[A^2-\left(\frac{A}{2}\right)^2\right] \\ & K=\frac{1}{2} m \omega^2\left(\frac{3 A^2}{4}\right)---(2)\end{aligned}$
From equation (1) and (2)
$\frac{K}{U}=\frac{3}{1}$
The kinetic energy of the particle at $t=\frac{T}{12}$
$\begin{aligned} & K=\frac{1}{2} m \omega^2 x^2\left(A^2-x^2\right)=\frac{1}{2} m \omega^2\left[A^2-\left(\frac{A}{2}\right)^2\right] \\ & K=\frac{1}{2} m \omega^2\left(\frac{3 A^2}{4}\right)---(2)\end{aligned}$
From equation (1) and (2)
$\frac{K}{U}=\frac{3}{1}$