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A particle of mass m is moving in a potential well, for which the potential energy is  given by U(x)=U01sinbx where  U0 and a are constants. Then for small  oscillations find time period

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a
time period, T=2πmbU0
b
speed of the particle is maximum at x = b
c
amplitude of oscillations is πb
d
time period, T=2πmb2U0

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detailed solution

Correct option is D

given PE is U(x)=U01−sinbx then force is F=−dUdx differentiate given equation wrt x F=−bU0sinbx  For small oscillations, sinbx≈bx  ⇒F=−b2U0x---(1) for simple harmonic motion F=-mω2x---(2) substitute eqn(2) in (1) -mω2x=−b2U0x ω=b2U0m=2πT T=2πmb2U0=time period of small oscillation here ω=angular velocity; m=mass ;x=displacement of oscillator .


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