Physics - SHM Question with Solution | TestHub

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

At any angle $\varphi$, the tension $T$ in the string is given by:

$T-mg\cos \varphi =\frac{m{v}^2}{l}$

By conservation of energy, the velocity $v$ at angle $\varphi$ is given by:

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

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

Substituting this into the tension equation, we get:

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

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

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

The minimum tension occurs at the extreme positions ($\varphi =\theta$):

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

Given that $T_{\max }=2T_{\min }$:

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

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

$4\cos \theta =3$

$\cos \theta =0.75$

$\theta ={\cos }^{-1}(0.75)$

Answer: ${\cos }^{-1}(0.75)$