Question

A particle starts oscillating simple harmonically from its equilibrium position. Then the ratio of kinetic and potential energy of the particle at time $\frac{T}{12}$ is
($T=$ time period and assume potential energy at equilibrium position$=0$)

• A. $2:1$
• B. $3:1$
• C. $4:1$
• D. $1:4$

Answer 1

The correct option is B $3:1$

Let, the equation of the particle performing SHM is given by $x=A\sin \omega t$
where $\omega =\frac{2\pi }{T}$
At $t=\frac{T}{12},x=A\sin \frac{2\pi }{T}.\frac{T}{12}=A\sin \frac{\pi }{6}=\frac{A}{2}$
The potential energy of the particle at $t=\frac{T}{12}$
$U=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{\left(\frac{A}{2}\right)}^2...\left(1\right)$
The kinetic energy of the particle at $t=\frac{T}{12}$
$K=\frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}m{\omega }^2[A^2-{\left(\frac{A}{2}\right)}^2]$
$K=\frac{1}{2}m{\omega }^2\left(\frac{3A^2}{4}\right)...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$\frac{K}{U}=\frac{3}{1}$