Question

The string of a simple pendulum is replaced by a uniform rod of length $L$ and mass $M$ while the bob has a mass $m$. It is allowed to make small oscillations. Its time period is?

• A. $2\pi \sqrt{\left(\frac{2M}{3m}\right)\frac{L}{g}}$
• B. $2\pi \sqrt{\frac{(M+3m)L}{3(M+2m)g}}$
• C. $2\pi \sqrt{\frac{(M+m)}{(M+3m)}\frac{L}{g}}$
• D. $2\pi \sqrt{\left(\frac{2m+M}{3(M+2m)}\right)\frac{L}{g}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{(M+3m)L}{3(M+2m)g}}$

Time period of Compound Pendulum is given by
$T=2\pi \sqrt{\frac{I_0}{Mgd}}$
$I_0=$ moment of inertia about point of suspension
$=\frac{M{L}^2}{3}+m{L}^2=\frac{(M+3m)L^2}{3}$
$M=$ total mass $=(M+m)$
$d=$ sepration between point of suspension and centre of mass of the pendulm
$=\frac{M\left(L/2\right)+m\left(L\right)}{(M+m)}=\frac{(M+2m)L}{2(M+m)}$
Substituting the values in Eq. (i), we get
$\Rightarrow T=2\pi \sqrt{\frac{2(M+3m)L}{3(M+2m)g}}$