Question

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

• A. 1 m/s
• B. 1.5 m/s
• C. 3 m/s
• D. 2 m/s

Answer 1

The correct option is B 1.5 m/s

At bottom velocity of bob is maximum$=A\omega$

Equation of motion, $x=A\sin \omega t$ when $\omega t={60}^{\circ }$

$x=\frac{\sqrt{3}}{2}A$

Speed of bob is $v=\omega \sqrt{(A^2-x^2)}=A\frac{\omega }{2}$

At $x=\frac{\sqrt{3}}{2}A$

here $A\omega =3$ we get,

$v=\frac{3}{2}=1.5m/s$