First slide

Applications of SHM

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

Difficult

A

$2 π \sqrt{(\frac{2 M}{3 m}) \frac{L}{g}}$
B

$2 π \sqrt{\frac{2 (M + 3 m) L}{3 (M + 2 m) g}}$
C

$2 π \sqrt{(\frac{M + m}{M + 3 m}) \frac{L}{g}}$
D

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

Solution

When the rod is replaced by the string the simple pendulum will act as compound pendulum for which time period is given by
$T = 2 π \sqrt{\frac{I_{0}}{M g d}}$ ………(i)
I₀= moment of inertia about point of suspension
$= \frac{M L^{2}}{3} + m L^{2} = \frac{(M + 3 m) L^{2}}{3}$
M' = total mass = (M + m)
d = separation between point of suspension and centre of mass of the pendulum
$= \frac{M (L / 2) + m (L)}{(M + m)} = \frac{(M + 2 m) L}{2 (M + m)}$
Substituting the values in Eq. (i), we get
$\Rightarrow T = 2 π \sqrt{\frac{2 (M + 3 m) L}{3 (M + 2 m) g}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App