The equation of motion of a particle executing simple harmonic motion is given by $x=3 \sin \left(6…
The equation of motion of a particle executing simple harmonic motion is given by $x=3 \sin \left(6 t+\frac{\pi}{6}\right)$, where $x$ is in metres and $t$ is in seconds. The ratio of the potential and kinetic energies of the particle at time $t=0$ is
$1: 1$
$1: 4$
$1: 2$
$1: 3$
Solution
Equation of motion is given as
$x=3 \sin \left(6 t+\frac{\pi}{6}\right), A=3$
At time $\mathrm{t}=0$
$x=3 \sin \left(\frac{\pi}{6}\right)=\frac{3}{2}=1.5$
Potential Energy of SHM is given as:
$\mathrm{V}=\frac{1}{2} \mathrm{kx}^2$ $\ldots$ (1)
Kinetic Energy of SHM is given as:
$K=\frac{1}{2} K\left(A^2-x^2\right)$ ...(2)
Dividing Equation (1) and (2)
$\begin{aligned} & \frac{V}{K}=\frac{1}{2} \frac{K x^2}{K\left(A^2-x^2\right)} \times 2 \\ & =\frac{x^2}{A^2-x^2}\end{aligned}$
Substitute, $x=1.5$ and $A=3$
$\begin{aligned} & \frac{\mathrm{V}}{\mathrm{K}}=\frac{(1.5)^2}{(3)^2-(1.5)^2}=\frac{2.25}{9-2.25} \\ & =\frac{225}{675}=\frac{45}{135}=\frac{9}{27}=\frac{1}{3} \\ & \frac{\mathrm{V}}{\mathrm{K}}=\frac{1}{3}\end{aligned}$