Question

The mass of the bob of a simple pendulum is ${}^{\prime }{m}^{\prime }$ and it carries a charge ${}^{\prime }{q}^{\prime }$. It oscillates in a uniform electric field $E$ as shown in the figure. Its time period will be

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

Answer 1

The correct option is B $2\pi {\left(\frac{l}{g+\frac{qE}{m}}\right)}^{\frac{1}{2}}$

Given:
Mass of bob of pendulum $=m$

Charge on bob = $+q$

Electric field intensity $=E$


Force applied by electric field on bob $=qE$

$F=qE$

$ma=qE$

$a=\frac{qE}{m}$

$g_{\text{eff}}=g-a$

$g_{\text{eff}}=g-\frac{qE}{m}$

Time period of oscillation, $T=2\pi \sqrt{\frac{l}{g_{\text{eff}}}}$

i.e. $T=2\pi {\left(\frac{l}{g-\frac{qE}{m}}\right)}^{\frac{1}{2}}$