Question

A simple pendulum of length $l$ and bob mass m is displaced from its equilibrium position O to a position P so that the height of P above O is $h$ and is 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)

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

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. 9.25). If V is the velocity of the bob at 0, then
$\frac{1}{2}m{V}^2=mgh$
or, $V^2=2gh$
At position 0, the tension F in the string is given by
F – mg = centripetal force = $\frac{m{V}^2}{l}$
F=mg+=$\frac{m{V}^2}{l}=mg+\frac{m{V}^2}{l}$ $\because V^2=2gh$
or, F=mg $(1+\frac{2h}{l})$
Hence the correct choices are (a) and (d)