Question

The bob of a simple pendulum of mass ${}^{\prime }{m}^{\prime }$ is performing oscillations such that the tension in the string is equal to twice the weight of the bob while it is crossing the mean position. The tension in the string when the bob reaches extreme position is :

• A. $\frac{mg}{2}$
• B. $mg$
• C. $\frac{3mg}{2}$
• D. zero

Answer 1

The correct option is A $\frac{mg}{2}$

At bottom $T=2mg$
$T-mg=\frac{m{V}^2}{R}$
$2mg-mg=\frac{m{V}^2}{R}$
$\sqrt{Rg}=V$
Applying Work Energy Theorem between pt. (1) and (2)
$mgR(1-cos\theta )=\frac{1}{2}m{V}^2$
$mgR(1-cos\theta )=\frac{mRg}{2}$

here, $\theta$ is the angle made by the particle with the axis at its extreme point,

$1-cos\theta =\frac{1}{2}$
$\theta ={60}^0$
at, $\theta ={60}^0$
$T=mgcos\theta$
$\therefore T=mg\frac{1}{2}$