Question

A bob is suspended from an ideal string of length $l$. Now it is pulled to a side through ${60}^o$ to vertical and rotates along a horizontal circle. Then its period of revolution is

• A. $2\pi \sqrt{l/g}$
• B. $\pi \sqrt{l/2g}$
• C. $\pi \sqrt{2l/g}$
• D. $\pi \sqrt{l/g}$

Answer 1

The correct option is C $\pi \sqrt{2l/g}$

Solution:

Given: Bob makes an angle of ${60}^{\circ }$ with vertical

As bob is whirled around horizontal circle, so a centrifugal force ($F_c$ ) act along with other force

Hence it is a non inertial frame, centrifugal force act as pseudo force

For vertical motion,

$T\cos {60}^{\circ }=mg$

$T=2mg$

For horizontal motion,

A string of length l makes an angle of ${60}^{\circ }$ with vertical, then the r of the circular motion of bob will be $(sin\theta =\frac{r}{l})$ $\Rightarrow lsin\theta$

$T\sin {60}^{\circ }=F_c$

$T\sin {60}^{\circ }=m{\omega }^{2}l\sin {60}^{\circ }$

$Tsin{60}^{\circ }=m{\omega }^{2}l\sin {60}^{\circ }$

$\omega =\sqrt{\frac{2g}{l}}$

Time period =$\frac{2\pi }{\omega }$

$T=2\pi \sqrt{\frac{l}{2g}}=\pi \sqrt{\frac{2l}{g}}$

Hence, the correct option is (c).