Question

A pendulum bob is in SHM. The velocity of the bob in the mean position is V. If now the amplitude of oscillation is doubled and the length of the pendulum is also doubled, the speed of the bob in the mean position would be

• A. $V$
• B. $\sqrt{2}$ V
  
• C. V/2
• D. 4V

Answer 1

Answer: (B)

$\sqrt{2}$ V

Velocity in S.H.M is: $v=A\omega$
As length is doubled $T\propto \sqrt{l}$
$\Rightarrow T_1=\sqrt{2}T$
$\Rightarrow \omega =\frac{2\pi }{T}$
$\therefore v_1=\left(2A\right)\frac{\omega }{\sqrt{2}}=\sqrt{2}A\omega =\sqrt{2}V$