Question

The bob of a simple pendulum is hanging vertically down from a fixed identical bob by means of a string of length $l$. If both bobs are charged with a charge $q$ each , time period of the pendulum is:

(ignore the radii of the bob)

• A. $2\pi \sqrt{\frac{1}{g+\left(\frac{q^2}{l^{2}m}\right)}}$
• B. $2\pi \sqrt{\frac{1}{g-\left(\frac{q^2}{l^{2}m}\right)}}$
• C. $2\pi \sqrt{\frac{l}{g}}$
• D. $2\pi \sqrt{\frac{1}{g+\left(\frac{q}{l^{2}m}\right)}}$

Answer 1

The correct option is C $2\pi \sqrt{\frac{l}{g}}$

The force due to bob always acts along the lengths.
$\therefore$ Torque due to it is always 0.
$\tau =0$
$\therefore$ Time period does not change $T=2\pi \sqrt{\frac{l}{g}}$