First slide

Simple hormonic motion

Question

Maximum acceleration of the bob of a simple pendulum oscillating with amplitude A is 6m/s² and that of another pendulum having same amplitude ‘A’ is 3 m/s². A third pendulum is formed by joining the strings of the pendulum and its bob is allowed to oscillate with same amplitude ‘A’. Then maximum acceleration of this bob is

Moderate

A

9 m/s²
B

2 m/s²
C

4.5 m/s²
D

12 m/s²

Solution

For a simple pendulum, maximum acceleration is given by, $a = ω^{2} A = {(\sqrt{\frac{g}{l}})}^{2} A$
$\Rightarrow l = \frac{gA}{a}$
$∴$ Length of the third pendulum $l_{3} = l_{1} + l_{2} = \frac{gA}{a_{1}} + \frac{gA}{a_{2}}$
$∴$ Maximum acceleration of the bob of a third pendulum is, $a_{3} = \frac{gA}{l_{3}} = \frac{gA}{\frac{gA}{a_{1}} + \frac{gA}{a_{2}}} = \frac{a_{1} a_{2}}{a_{1} + a_{2}}$
$\Rightarrow a_{3} = \frac{6 \times 3}{6 + 3} m / s^{2} = 2 m / s^{2}$

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