Question

Mass of the bob of a simple pendulum of length $L$ is $m$. If the bob is projected horizontally from its mean position with velocity $\sqrt{4gL}$ , then the tension in the string becomes zero after a vertical displacement of :

• A. $L/3$
• B. $3L/4$
• C. $4L/3$
• D. $5L/3$

Answer 1

The correct option is D $5L/3$

Let the angle $\theta$ be when tension becomes zero.

Under this condition $mgsin\theta =\frac{m{v}^2}{L}$ ....... eq(1)

Also, by conservation of energy:

$\frac{m(\sqrt{4gL}{)}^2}{2}=mg(L+Lsin\theta )+\frac{m{v}^2}{2}$

=> $\frac{m{v}^2}{L}=2gmL-2gmLsin\theta$ ------- eq(2)

From eq (1) and eq(2):

$mgsin\theta =2mg-2mgsin\theta$

$sin\theta =2/3$

So, the tension in the string becomes zero after a vertical displacement of: $L+Lsin\theta =5L/3$