Question

The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.

Answer 1

The tension in the pendulum is maximum at the mean position and minimum on the extreme position.

Here $\left(1/2\right)m{v}^2-0=mgl(1-\cos \theta )$

$v^2=2gl(1-\cos \theta )$

Now, $T^{}=mg+2mg(1-\cos \theta )$ $[T=mg+(m{v}^2/l\left)\right]$

Again, $T^{}=mg\cos \theta$

According to question, $T_{max}=2T_{min}$

$\Rightarrow mg+2mg-2mg\cos \theta =2mg\cos \theta$

$\Rightarrow 3mg=4mg\cos \theta$

$\Rightarrow \cos \theta =3/4$

$\Rightarrow \theta ={\cos }^{-1}\left(3/4\right)$