Question

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of $60^{o}$ to the vertical, will be (If $g = 10m/s^{2}$)

• A. $\frac{E}{4}$
• B. $\frac{3E}{4}$
• C. $\frac{\sqrt{3}}{4}E$
• D. $K_{1}$

Answer 1

Answer: (D)

$K_{1}$

Answer 2

Let bob velocity be v at point B where it makes an angle of 60^o with the vertical, then using conservation of mechanical energy

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

⇒ $\frac{1}{2}m \times 3^{2} = \frac{1}{2}mv^{2} + mgl(1 - \cos\theta)$

$\therefore\frac{T_{2}}{T_{1}} = \sqrt{\frac{M_{2}}{M_{1}}} = \sqrt{\frac{4M}{M}} = 2$