Q1 / 214 · NTA Abhyas (NEET) 2020 · Physics · Oscillations

A particle of mass m is located in a field such that its potential energy is given by Ux=U01-cos⁡ax where U0 and a are positive constants. The period of small oscillations is

  • A
    2πU0ma2
  • B
    2πmU0a2
  • C
    2πa0mU0
  • Correct answer
    D
    2πmU0a2

Solution

Given potential energy is

Ux=U01-cos⁡ax

Therefore force on the particle is given by

F=-dUdx=-U0asin(ax)

Using small angle approximation, sin(ax)≈ax

Therefore F=-U0a2x

Time period of SHM is,

T=2πmk where k is force constant.

Therefore time period for given force is,

T=2πmU0a2

Hence, the correct answer is option: 'D'.

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