Question

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