Question

The angular amplitude of a simple pendulum is $ \theta_{0} $ The maximum-tension in its string will be

• A. $mg(1 - \theta_0)$
• B. $m g\left(1+\theta_{0}\right)$
• C. $m g\left(1-\theta_{0}^{2}\right)$
• D. $m g\left(1+\theta_{0}^{2}\right)$

Answer 1

Answer: (D)

The simple pendulum at angular amplitude $ {{\theta }_{0}} $ is shown in the figure.
$ {{T}_{\max }}=mg+\frac{m{{v}^{2}}}{l} \,\,\,...$ (i)
When bob of pendulum comes from $ A $ to $ B $ , it covers a vertical distance h.

$ \therefore \cos {{\theta }_{0}}=\frac{l-h}{l} $
$ h=l(1-\cos {{\theta }_{0}}) $ ... (ii)
Also during $ A $ to $ B $ , potential energy of bob converts into kinetic energy ie, $ mgh=\frac{1}{2}m{{v}^{2}} $
$ \therefore $ $ v=\sqrt{2gh} $ ... (iii)
Thus, using Eqs. (i), (ii) and (iii), we obtain
$ {{T}_{\max }}=mg+\frac{2mg}{l}l(1-\cos {{\theta }_{0}}) $
$ =mg+2mg\left[ 1-1+\frac{\theta _{0}^{2}}{2} \right] $
$=m g\left(1+\theta_{0}^{2}\right)$