Question

The bob of a pendulum of mass ($m$) at rest is given a sharp hit to impart the horizontal velocity $\sqrt{6gl}$, where $l$ is the length of the pendulum.

Find the tension in the string at highest point?

• A. $6mg$
• B. $3mg$
• C. $2mg$
• D. $mg$

Answer 1

The correct option is D $mg$

To reach highest point, bob moves distance $2l$ (vertically).

$\Rightarrow$Since $W_T=0$; because at each point tension ($T$) is $\perp$ to displacement, so ($W_T=T.ds=0)$ and gravity is a conservative force, using energy conservation between bottom position and top position of the bob:

$K{E}_f+P{E}_f=K{E}_i+P{E}_i$
Assigning $PE=0$ at bottom position
$\Rightarrow \frac{1}{2}m{v}^2+mg\left(2l\right)=\frac{1}{2}m{u}^2+0$

$\therefore \frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2-2mgl$

Or $\frac{1}{2}m{v}^2=\frac{1}{2}m\left(6gl\right)-2mgl$

Or $\frac{1}{2}m{v}^2=mgl$

$\therefore v=\sqrt{2gl}$
applying equation of dynamics towards centre of circle, at the top position of bob


$\Rightarrow T+mg=\frac{m{v}^2}{l}$
$\therefore T+mg=\frac{m(\sqrt{2}gl{)}^2}{l}$
$\Rightarrow T+mg=2mg$
$\therefore T=mg$, at the topmost position of the bob.