A particle of mass m is moving such that its potential energy is given by U(x)=U0(1-Cosax) where U0 and A are constant. The time period of small oscillation is

Dear Student ,
Here in this case from the given question we can write that ,
We know that ,F=-dUdxSo , Fx=-U0a sinaxNow for small oscillation sinax=ax So , F=-U0a2xSo by the equation of motion we get ,md2xdt2+U0a2x=0So , ω02=U0a2mTherefore , time period ,T=2πω0=2πamU0

Hope this helps you .
Regards

  • -18
What are you looking for?