A simple pendulum has a length l, mass of bob m. The bob is given a charge q. The pendulum is suspended between the vertical plates of the charged parallel plate capacitor. If E is the field strength between the plates, then time period T equal to:
A
2π√lg
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B
2π
⎷lg+qEm
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C
2π
⎷lg−qEm
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D
2π
⎷l√g2+(qEm)2
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Solution
The correct option is D2π
⎷l√g2+(qEm)2 Here two force is applied - one is gravitational force (mg) and other is electrostatic force on the pendulum bob. The resultant force , →F=mg^j+qE^i. If aeff is the effective acceleration of pendulum bob , m→aeff=→F=mg^j+qE^i. ∴|→aeff|=aeff=√g2+(qEm)2 Now time period , T=2π√laeff=2π
⎷l√g2+(qEm)2