A particle of mass m moves on a smooth horizontal line AB of length a. At a point P, a force mg(AP)/a acts towards A and another force 2mg(BP)/a acts towards B.
Physics · Simple Harmonic Motion
JEE Advanced

Question

A particle of mass m moves on a smooth horizontal line AB of length a. At a point P, a force mg(AP)/a acts towards A and another force 2mg(BP)/a acts towards B.

G1 Physics · Accepted Answer

Let x=AP measured from A, so BP=a-x. Net force toward +B is 2mg(a-x)/a-mgx/a=2mg-3mgx/a=-(3mg/a)(x-2a/3). Thus equilibrium is x₀=2a/3 and ω²=3g/a, so T=2π√(a/3g): Sub 1=D. Released from x=a/2, amplitude is a/6 and nearest B-side extreme is x=5a/6. After the A-directed force is removed there, work by the remaining B-directed force to x=a is the integral of 2mg(a-x)/a from x=5a/6 to x=a=mga/36. Hence (1/2)mv²=mga/36 and v=√(2ga)/6: Sub 2=B.