Question

A simple pendulum of string length l and bob of mass m is displaced from its equilibrium position O to a position P so that the height of P above O is h. It is then released. What is the tension in the string when the bob passes through the equilibrium position O? Neglect friction. V is the velocity of the bob at O.

• A. $m(g+\frac{v^2}{l})$
• B. $\frac{2mgh}{l}$
• C. $mg(1+\frac{h}{l})$
• D. $mg(1+\frac{2h}{l})$

Answer 1

Answer: (A), (D)

The correct options are
A

$m(g+\frac{v^2}{l})$

D

$mg(1+\frac{2h}{l})$

P.E. at point P = mgh. If friction is neglected, the potential is completely converted into kinetic energy when the bob reaches the equilibrium position O (see fig). If V is the velocity of the bob at O, then

$\frac{1}{2}m{V}^2=mgh$

or $V^2=2gh$

At position O, the tension F in the string is given by
F – mg = centripetal force

$=\frac{m{V}^2}{1}$

or $F=mg+\frac{m{V}^2}{1}$

$=mg+\frac{2mgh}{l}$

or $F=mg(1+\frac{2h}{l})$

Hence the correct choices are (a) and (d)