Question

A particle of mass 10 gm is placed in a potential field given by $V=(50x^2+100)J/kg$. The frequency of oscillation in cycles/sec is :

• A. $\frac{10}{\pi }$
• B. $\frac{5}{\pi }$
• C. $\frac{100}{\pi }$
• D. $\frac{50}{\pi }$

Answer 1

Answer: (B)

$\frac{5}{\pi }$

Potential energy $U=mV$
$\Rightarrow U=(50x^2+100){10}^{-2}$
$F=-\frac{dU}{dx}=-\left(100x\right){10}^{-2}$
$\Rightarrow m{\omega }^{2}x=-(100\times {10}^{-2})x$
$10\times {10}^{-3}{\omega }^{2}x=100\times {10}^{-2}x$
$\Rightarrow {\omega }^2=100,\omega =10$
$\Rightarrow f=\frac{\omega }{2\pi }=\frac{10}{2\pi }=\frac{5}{\pi }$