Physics Circle home Student Login

Home›Chapters›Work, Energy and Power PYQs›Potential Energy & Equilibrium›Question 13

NEET 2011 (AIPMT, Prelims) · Potential Energy & Equilibrium

Potential Energy & Equilibrium question from NEET 2011 (AIPMT, Prelims)

Q13NEET 2011 · AIPMT · PRELIMSMedium

Potential Energy & Equilibrium

A particle of mass m is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time?

Figure

Show answer and solution

Answer: (b)

  1. V∝x2V \propto x^2 gives F=−kxF = -kx, so the motion is SHM about x=0x = 0.
  2. Released from rest at a positive xx, so x=Acos⁡ωtx = A\cos\omega t: starts at maximum, smooth curve.
  3. So the answer is (b).
Report a mistake

More questions on Potential Energy & Equilibrium