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 :

Question

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Answer 1

The bob of pendulum executing $S . H . M$ has maximum velocity at the mean position.
Maximum velocity is given by $v_{m a x} = A w$
where $A$ is the amplitude of oscillation and $w$ is the angular frequency given by $w = \frac{g}{l}$
$∴$ For the same length of the string, $w$ remains the same.
$⟹ v_{m a x} \propto A$
Thus $A^{'} = 2 A ⟹ v_{m a x}^{'} = 2 v$

Answer 2

The bob of pendulum executing

S . H . M

has maximum velocity at the mean position.

Maximum velocity is given by

v_{m a x} = A w

where

A

is the amplitude of oscillation and

w

is the angular frequency given by

w = \frac{g}{l}

∴

For the same length of the string,

w

remains the same.

⟹ v_{m a x} \propto A

Thus

A^{'} = 2 A ⟹ v_{m a x}^{'} = 2 v