Questions
A particle of mass m is moving in a potential well, for which the potential energy is given by where and a are constants. Then for small oscillations find time period
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 .Similar Questions
A body executes S.H.M. under the action of a force F1 with a time period 7/6 seconds. If the force is changed to F2 it executes S.H.M. with time period 7/8 seconds. If both the forces F1 and F2 act simultaneously in the same direction on the body, then its time period in seconds is
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