Question

The maximum tension in the string of a pendulum is two times the minimum tension. Let ${\theta }_0$ be the angular amplitude ,Then ${\theta }_0$ is :

• A. 1/2
• B. 3/4
• C. 2/3
• D. 3/5

Answer 1

Answer: (B)

3/4

$\begin{array}{c}\text{Tension max at Mean}\\ T_{\text{max}}=mg+\frac{m{v}^2}{l}.....\left(1\right)\end{array}$

$\begin{array}{c}\text{Tension min at extreme.}\\ T_{\text{min}}=\mathrm{mg}\cos \theta .....\left(ii\right)\\ \text{According to question}\end{array}$

$\begin{array}{c}{T}_{max}=2T_{min}\\ mg+\frac{m{v}^2}{l}=2mg\cos {\theta }_0\\ g+\frac{v^2}{l}=g\cos {\theta }_0\end{array}$

$1+\frac{v^2}{g^2}=\cos \theta ......\left(iii\right)$

$\begin{array}{c}\text{Also, conservation of}\\ \text{mechanical energy}\\ \frac{1}{2}m{v}^2=\mathrm{mgl}(1-\cos \theta )\end{array}$

$\begin{array}{c}{v}^2=2gl(1-\cos \theta ).....\left(iv\right)\\ \text{From}\\ \text{(iv)}and\text{(ii)}\end{array}$

$\begin{array}{c}1+2(1-\cos \theta )=2\cos \theta \\ 1+2-2\cos \theta =2\cos \theta \\ 3=4\cos \theta \\ \cos \theta =\frac{3}{4}\Rightarrow \theta ={53}^{\circ }\\ \theta ={\cos }^{-1}3/4\end{array}$