Question

The time for a particle executing SHM with frequency 1 Hz is considered from initial phase of $\frac{\pi }{4}$. The time at which potential energy is three times of kinetic energy is (in second):

• A. $\frac{1}{12}$
• B. $\frac{1}{24}$
• C. $\frac{1}{6}$
• D. $\frac{1}{48}$

Answer 1

The correct option is B

$\frac{1}{24}$

$x=Asin(\omega t+\varphi )$

$v=A\omega cos(\omega t+\varphi )$

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^{2}co{s}^2(\omega t+\varphi )$

$PE=\frac{1}{2}m{A}^2{\omega }^{2}si{n}^2(\omega t+\varphi )$

Given, PE = 3KE

So, $ta{n}^2(\omega t+\varphi )=3or\omega t+\varphi =\frac{\pi }{3}$

$\omega t=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}(given\varphi =\frac{\pi }{4})$

$2\pi ft=\frac{\pi }{12}$

$t=\frac{1}{24}$ seconds.