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A simple pendulum of length L is placed between theplates of a parallal plate capacitor having electric field E, as shown in figure. its bob has mass m and charge q. The time period of the pendulum is given by:

Option 1)
$2\pi \sqrt{\frac{L}{\sqrt{g-\frac{qE}{m}}}}$

Option 2)
$2\pi \sqrt{\frac{L}{\sqrt{g^{2}-\frac{q^{2}E^{2}}{m^{2}}}}}$

Option 3)
$2\pi \sqrt{\frac{L}{g+\frac{qE}{m}}}$
Option 4)
$2\pi \sqrt{\frac{L}{\sqrt{g^{2}+\left ( \frac{qE}{m} \right )^{2}}}}$

Answers (1)

A Aadil

Two force act on the mass m

1) gravity force

2) force by electric field.

net force on m is = $\sqrt{\left ( mg \right )^{2}+\left ( qE \right )^{2}}$

So, net g is = $\sqrt{ g ^{2}+\left ( qE \right/m )^{2}}$

$T=2\pi \sqrt{\frac{l}{g}}$

$=2\pi \sqrt{\frac{l}{\sqrt{g^{2}+\left ( \frac{qt}{m} \right )^{2}}}}$

Option 1)

$2\pi \sqrt{\frac{L}{\sqrt{g-\frac{qE}{m}}}}$

Option 2)

$2\pi \sqrt{\frac{L}{\sqrt{g^{2}-\frac{q^{2}E^{2}}{m^{2}}}}}$

Option 3)

$2\pi \sqrt{\frac{L}{g+\frac{qE}{m}}}$

Option 4)
$2\pi \sqrt{\frac{L}{\sqrt{g^{2}+\left ( \frac{qE}{m} \right )^{2}}}}$

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