Question

Displacement time graph of particle performing $SHM$ is shown in figure. Assume that mean position is at $x=0$, then kinetic energy and potential energy will be equal at:
[$T$ is time period and $A$ is amplitude]

• A. $t=\frac{T}{4}$
• B. $t=\frac{T}{8}$
• C. $t=\frac{T}{6}$
• D. $t=\frac{T}{12}$

Answer 1

The correct option is B $t=\frac{T}{8}$

At $t=0$, the displacement of particle is maximum and it is at the positive extreme position.
$x=+A$
(maximum displacement)
The equation of motion for partice will be,
$x=A\cos \omega t...\left(i\right)$
Now, $\text{kinetic energy}=\text{potential energy}$
$\Rightarrow \frac{1}{2}k(A^2-x^2)=\frac{1}{2}k{x}^2$
$\Rightarrow 2x^2=A^2$
$\Rightarrow x=\pm \frac{A}{\sqrt{2}}$
Since particle is at its positive extreme, the $P.E$ and $K.E$ will attain equal values for position of particle between its positive extreme and the mean position.
$\Rightarrow x=+\frac{A}{\sqrt{2}}...\left(ii\right)$
From Eq.$\left(i\right)$ and Eq.$\left(ii\right)$,
$A\cos \omega t=+\frac{A}{\sqrt{2}}$
Or, $\cos \omega t=+\frac{1}{\sqrt{2}}$
$\Rightarrow \omega t=\frac{\pi }{4}\Rightarrow \frac{2\pi }{T}\times t=\frac{\pi }{4}$
$\therefore t=\frac{T}{8}$