JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Elasticity & SHM

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A simple pendulum of length L is placed between the plates of a parallel 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

(A) $2 π \frac{L}{g ^{2} + ( \frac{q E}{m} ) ^{2}}$ ✓ Correct
(B) $2 π \frac{L}{( g + \frac{q E}{m} )}$
(C) $2 π \frac{L}{( g - \frac{q E}{m} )}$
(D) $2 π \frac{L}{g ^{2} - \frac{q ^{2} E ^{2}}{m ^{2}}}$

Answer: (A)

Solution

The bob feels gravity

m g

pulling down and the electric force

q E

pushing sideways, so it hangs along an effective gravity of magnitude

g_{eff} = g^{2} + (\frac{q E}{m})^{2}

. The pendulum's period uses this effective value in place of

g

, giving

T = 2 π \frac{L}{g ^{2} + ( \frac{q E}{m} ) ^{2}}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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