Question

A pendulum is going S.H.M. The velocity of the bob in the mean position is $v$. If now its amplitude is doubled, keeping the length same, its velocity in the mean position will be :

• A. $v/2$
• B. $v$
• C. $2v$
• D. $4v$

Answer 1

Answer: (C)

$2v$

The bob of pendulum executing $S.H.M$ has maximum velocity at the mean position.

Maximum velocity is given by $v_{max}=Aw$

where $A$ is the amplitude of oscillation and $w$ is the angular frequency given by $w=\sqrt{\frac{g}{l}}$

$\therefore$ For the same length of the string, $w$ remains the same.

$⟹v_{max}\propto A$

Thus $A^{\prime }=2A⟹v_{max}^{\prime }=2v$