A simple pendulum is suspended from the roof of a trolley which moves, in a horizontal direction with an acceleration a, and its time period is given by T=2π√lg′ which of the following correctly represent the value of g′.
A
g
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B
g−a
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C
g+a
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D
√g2+a2
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Solution
The correct option is D√g2+a2 Since, the trolley is moving with an acceleration ′a′, a pseudo force (ma) will act on the simple pendulum, when observed from frame of trolley
Hence net force on bob is, Fnet=m√g2+a2 For effective value of g, ⇒mg′=m√g2+a2 mg′ is the effective weight of bob. ∴g′=√g2+a2 ⇒geff=g′ is equal to √g2+a2, will be used in finding time period of pendulum. T=2π√lg′