Question

The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above Q and the bob is then released. Assuming the mass of the bob to be m and time period of oscillation to be 2.0 seconds, the tension in the string when the bob passes through O is

• A. $m(g+2{\pi }^{2}h)$
• B. $m(g+{\pi }^{2}h)$
• C. $m(g+\frac{{\pi }^2}{2}h)$
• D. $m(g+\frac{{\pi }^2}{3}h)$

Answer 1

Answer: (A)

$m(g+2{\pi }^{2}h)$

When the bob is at O, the tension in the string is $T=mg+\frac{m{v}^2}{l}$

where $mgh=\frac{1}{2}m{v}^2$ or $v{=}^2=2gh$
$\therefore T=mg+\frac{2mgh}{l}$
here time period $=2\pi \sqrt{l/g}=2$ or $\frac{1}{{\pi }^2}=\frac{l}{g}$
Thus, $T=m[g+2{\pi }^{2}h]$