Question

A pendulum is made from a massless rod of length L and bob of mass m. A spring of force constant k is connected horizontally to it at a distance h below its point of suspension as shown. The rod is in equilibrium in vertical position. The frequency of vibration of the system for small values of $\theta$ is :

• A. $\frac{1}{2\pi L}\sqrt{gL+\frac{kh}{m}}$
• B. $\frac{1}{2\pi L}\sqrt{\frac{mgL+k}{m}}$
• C. $2\pi \sqrt{\frac{m{L}^2}{mgL+kh}}$
• D. $\frac{1}{2\pi L}\sqrt{gL+\left(\frac{k{h}^2}{m}\right)}$

Answer 1

The correct option is D

$\frac{1}{2\pi L}\sqrt{gL+\left(\frac{k{h}^2}{m}\right)}$

The extension developed in the string due to small values of '$\theta$' is :

$x=hsin\theta \cong h\theta$

torque about 'O' :

${\tau }_0=\left(mgsin\theta \right)L+\left(kx\right)h$

or, ${\tau }_0\cong mg\theta L+kh2\theta =(mgL+k{h}^2)\theta \dots \left(1\right)$

Also,

${\tau }_0=I_0\alpha =m{L}^2\alpha \dots \left(2\right)$

From (1) and (2) :

$m{L}^2\alpha =(mgL+k{h}^2)\theta$

or $\alpha =\frac{1}{L^2}(gL+\frac{k{h}^2}{m})$

Now : $T=2\pi \sqrt{\frac{\theta }{\alpha }}=2\pi \sqrt{\frac{\theta }{\frac{1}{L^2}(gL+\frac{k{h}^2}{m})\theta }}$

$\Rightarrow f=\frac{1}{T}=\frac{1}{2\pi L}\sqrt{gL+\left(\frac{k{h}^2}{m}\right)}$

Hence (d)