Questions
A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is
detailed solution
Correct option is A
U=k|x|3⇒F=−dUdx=−3k|x|2 ........(i) Also, for SHM, x=asinωt and d2xdt2+ω2x=0⇒ Acceleration a=d2xdt2=−ω2x⇒F=ma=md2xdt2=−mω2x .......(ii) From eqs. (i) and (ii), we get ω=3kxm⇒T=2πω=2πm3kx=2πm3k(asinωt)⇒T∝1aSimilar Questions
When the displacement is half the amplitude, the ratio of potential energy to the total energy is
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