Question

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

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

Answer 1

The correct option is D $\frac{5}{\pi }$

Given,
Mass of the particle
$m=10g$
And potential feild,
$V=(50x^2+100)J/kg$
Potential energy,
$U=mV\Rightarrow =10\times {10}^{-3}\times (50x^2+100)$
Force can be written as rate of change of potential with respect to distance travelled,
$F=-\frac{dU}{dx}\Rightarrow F=-\frac{d\left(\right(50x^2+100)\times 10\times {10}^{-3})}{dx}\Rightarrow F=-(1000x\times {10}^{-3})=-x$
As we know,
$F=-kx=-m{\omega }^{2}x\Rightarrow -m{\omega }^{2}x=-x\Rightarrow 10\times {10}^{-3}{\omega }^{2}x=x\Rightarrow {\omega }^2=100\Rightarrow \omega =10$
The frequency of oscillation,
$f=\frac{\omega }{2\pi }=\frac{10}{2\pi }\Rightarrow f=\frac{5}{\pi }$