Question

The time period of SHM performed by the simple pendulum -

• A. T = 2p$\sqrt{\mathcal{l/}g}$
• B. T = 2p$\sqrt{\frac{\mathcal{l}}{g - qE/m}}$
• C. T = 2p$\sqrt{\frac{\mathcal{l}}{\sqrt{g^{2} + (qE/m)^{2}}}}$
• D. None

Answer 1

Answer: (C)

T = 2p$\sqrt{\frac{\mathcal{l}}{\sqrt{g^{2} + (qE/m)^{2}}}}$

Answer 2

T = 2p