Question

If a simple pendulum oscillates with an amplitude of $50\text{mm}$ and time period of $2\text{seconds}$, then its maximum velocity is

• A. $0.8\text{m/s}$
• B. $0.5\text{m/s}$
• C. $2.1\text{m/s}$
• D. $0.16\text{m/s}$

Answer 1

The correct option is D $0.16\text{m/s}$

Given : Time period $T=2\text{seconds}$; amplitude of pendulum $A=50\text{mm}=0.05\text{m}$
We know that the velocity of a simple pendulum undergoing SHM is given by

$\therefore v_{max}=\frac{2\pi }{T}A$

$\therefore v_{max}=\frac{2\pi }{2}\times 0.05=0.16\text{m/s}$

Hence, the correct answer is option (d).

Why this question? Note: In simple harmonic motion, maximum velocity occurs at mean position.