Question

If a simple pendulum oscillates with an amplitude of $50mm$ and time period of $2sec$, then its maximum velocity, is

• A. 0.8 m/s
• B. 0.15 m/s
• C. 0.10 m/s
• D. 0.16 m/s

Answer 1

Answer: (D)

Given : time period $T=2sec$
amplitude of pendulum $A=50mm=0.05m$ We know that the velocity of a simple pendulum undergoing SHM is given by
$v=\omega \sqrt{A^2-x^2}=\frac{2\pi }{T}\sqrt{A^2-x^2}$
$\therefore v_{m2x}=\frac{2\pi }{T}\sqrt{A^2-0^2}$($\because$ maximum velocity
occurs at $x=0$ )
$\therefore v_{\max }=\frac{2\pi }{T}\times 0.05$
$=0.16m/s$