Physics - NLM Question with Solution | TestHub

For the mass $M$ to remain stationary, the tension $T$ in the string must balance its weight:

$T=Mg$

For the mass $m$ revolving in a horizontal circle of radius $L$, the necessary centripetal force is provided by the tension $T$ in the string. If $\omega$ is the angular velocity of mass $m$, then:

$T=m{\omega }^{2}L$

Equating the two expressions for tension, we get:

$Mg=m{\omega }^{2}L$

Since angular velocity $\omega =2\pi f$, where $f$ is the frequency of revolution, we can substitute this into the equation:

$Mg=m(2\pi f{)}^{2}L$

$Mg=4{\pi }^{2}m{f}^{2}L$

Solving for $f^2$:

$f^2=\frac{Mg}{4{\pi }^{2}mL}$

Taking the square root on both sides, we get the frequency:

$f=\frac{1}{2\pi }\sqrt{\frac{Mg}{mL}}$

Answer: $\frac{1}{2\pi }\sqrt{\frac{\mathrm{Mg}}{\mathrm{mL}}}$