question_answer

A simple pendulum has a length I and the mass of the bob is m. The bob is given a charge q coulomb. The pendulum is suspended between the vertical plates of a charged parallel plate capacitor. If E is the electric field strength between the plates, the time period of the pendulum is given by:

A)  \[2\pi \sqrt{\frac{l}{g}}\]

B)  \[2\pi \sqrt{\frac{l}{\sqrt{g+\frac{qE}{m}}}}\]

C)  \[2\pi \sqrt{\frac{l}{\sqrt{g-\frac{qE}{m}}}}\]

D) \[2\pi \sqrt{\frac{l}{\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}}}\]

Correct Answer: D

Solution :

Time period of simple pendulum in air \[T=2\pi \sqrt{\frac{l}{g}}\] When it is suspended between vertical plates of a charged parallel plate capacitor, then accelertion due to electric field,\[a=\frac{qE}{m}\] This acceleration is acting horizontally and acceleration due to gravity is acting vertically. So, effective acceleration \[g=\sqrt{{{g}^{2}}+{{a}^{2}}}=\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}\] Hence,\[T=2\pi \sqrt{\frac{l}{\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}}}\]