The bob of simple pendulum is hanging vertically down from a fixed identical bob by means of a sting of…

The bob of simple pendulum is hanging vertically down from a fixed identical bob by means of a sting of length $l$. If both bobs are charged with a charge $q$ each, time period of the pendulum is : (ignore the radii of the bobs)
  1. $2 \pi \sqrt{\frac{l}{g+\left(\frac{q^2}{l^2 m}\right)}}$
  2. $2 \pi \sqrt{\frac{l}{g-\left(\frac{q^2}{l^2 m}\right)}}$
  3. $2 \pi \sqrt{\frac{l}{g}}$
  4. $2 \pi \sqrt{\frac{l}{g-\left(\frac{q^2}{l}\right)}}$

Solution

There will be a electrostatic repulsion between two charged bobs, but it does not affect the motion of pendulum. Thus, time period of pendulum remains same i.e., $T=2 \pi \sqrt{\frac{l}{g}}$.

Asked in: AP EAMCET 2006

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