Question

A particle of mass $10\, gm$ is placed in a potential field given by $V=\left(50 x^{2}+100\right) \, J / kg$. The frequency of oscillation in cycle/sec is

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

Answer 1

Answer: (B)

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