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 :

$x=htan\theta \cong h\theta$

torque about 'O' :

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

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

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

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

Hence (d)