The string of a simple pendulum is replaced by a uniform rod of l… - Vidyadip

MCQ

The string of a simple pendulum is replaced by a uniform rod of length $L$ and mass $M$. If the mass of the bob of the pendulum is $m$, then for small oscillations its time period would be (assume radius of bob $r << L$)

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

✓

Answer

Correct option: A.

$2\pi \sqrt {\frac{{2(M + 3m)\,L}}{{3(M + 2m)\,g}}} $

a
When the rod is replaced by the string the simple pendulum will act as compound pendulum for which time period is given by

$T=2 \pi \sqrt{\frac{I_{0}}{M g d}}$ $...(i)$

$I_{0}=$ moment of inertia about of suspension

$=\frac{M L^{2}}{3}+m L^{2}=\frac{(M+3 m) L^{2}}{3}$

$M^{\prime}=$ total mass $=(M+m)$

$d=$ separation between point of suspension and centre of mass of the pendulum

$=\frac{M\left(\frac{L}{2}\right)+m(L)}{(M+m)}=\frac{(M+2 m) L}{2(M+m)}$

Substituting the values in Eq. $(i)$ we get

$\Rightarrow T=2 \pi \sqrt{\frac{2(M+3 m) L}{3(M+2 m) g}}$

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