Question

The angular amplitude of a simple pendulum is ${θ}_0$. The maximum tension in its string will be

• A. $mg\left(1−{θ}_0\right)$
• B. $mg\left(1+{θ}_0\right)$
• C. $mg\left(1−{θ}_0^2\right)$
• D. $mg\left(1+{θ}_0^2\right)$

Answer 1

Answer: (D)

The simple pendulum at angular amplitude ${θ}_n$ is shown in figure.
Maximum tension in the string is
$T_{\max }=mg+\frac{m{v}^2}{l}…(i)$
When bob of the pendulum comes form $A$ to $B$, it covers a vertical distance $h$
$∴\cos θ=\frac{l−h}{h}$
$⇒h=l\left(1−\cos {θ}_0\right)…(ii)$
Also, during $B$ to $A$, potential energy of bob converts into kinetic energy, i.e.
$mgh=\frac{1}{2}m{v}^2$
$∴v=\sqrt{2gh}…(iii)$
Thus, using Eqs. (i), (ii) and (iii), we obtain
$T_{\max }=mg+\frac{2mg}{l}l\left(1−\cos {θ}_0\right)$
$=mg+2mg\left[1−1+\frac{{θ}_0^2}{2}\right]$
$=mg\left(1+{θ}_0^2\right)$