Question

Find out time period of small oscillations from mean position

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

Answer 1

Answer: (C)

$2\pi \sqrt{\frac{l}{{(g^2+{\left(\frac{qE}{m}\right)}^2)}^{1/2}}}$

From FBD,
$Tsin\theta =qE$ ............... (1) where $T$ is the tension in the string.
$Tcos\theta =mg$ ............... (2)
$\therefore tan\theta =\frac{qE}{mg}$
$\theta =ta{n}^{-1}\left(\frac{qE}{mg}\right)$

Doing $(1{)}^2+(2{)}^2,$we will get,

$T^{2}si{n}^2\theta +T^{2}co{s}^2\theta =(mg{)}^2+(qE{)}^2$
$T=\sqrt{(mg{)}^2+(qE{)}^2}$
Speed of pendulum bob is $v=\sqrt{\frac{tension}{mass/length}}=\sqrt{\frac{T}{m/l}}$
Time period of pendulum $t=\frac{2\pi l}{v}=2\pi \sqrt{\frac{l}{T/m}}=2\pi \sqrt{\frac{l}{\left(\sqrt{(mg{)}^2+(qE{)}^2}\right)/m}}=2\pi \sqrt{\frac{l}{\left(\sqrt{[g^2+(qE/m{)}^2}\right]}}$
Ans:(C)