Physics - W.E.P. Question with Solution | TestHub

Let the mass of the bob be $m$ and the length of the pendulum be $l$.

The tension in the string at an angle $\alpha$ with the vertical is given by:

$T=mg\cos \alpha +\frac{m{v}^2}{l}$

By conservation of energy, the velocity $v$ at angle $\alpha$ is:

$\frac{1}{2}m{v}^2=mgl(\cos \alpha -\cos \theta )$

$\frac{m{v}^2}{l}=2mg(\cos \alpha -\cos \theta )$

Substituting this in the tension equation, we get:

$T=mg\cos \alpha +2mg(\cos \alpha -\cos \theta )=mg(3\cos \alpha -2\cos \theta )$

The maximum tension occurs at the lowest point ($\alpha =0$):

$T_{max}=mg(3-2\cos \theta )$

The minimum tension occurs at the extreme position ($\alpha =\theta$):

$T_{min}=mg(3\cos \theta -2\cos \theta )=mg\cos \theta$

Given that $T_{max}=4T_{min}$:

$mg(3-2\cos \theta )=4mg\cos \theta$

$3-2\cos \theta =4\cos \theta$

$6\cos \theta =3$

$\cos \theta =0.5$

$\theta ={\cos }^{-1}(0.5)$

Answer: ${\cos }^{-1}(0.5)$