question_answer

A simple pendulum has a length \[l\] 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) \[\sqrt{\frac{l}{\sqrt{g+\frac{qE}{m}}}}\]

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

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

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 acceleration 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,   \[T'=2\pi \]\[\sqrt{\frac{l}{\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}}}\]