The potential energy of a simple harmonic oscillator of mass 2 ~kg at its mean position is 5 ~J . If its total energy is 9 ~J and amplitude is 1 ~cm , then its time period is

Step-by-step solution

Given, total energy =9 ~J PE at mean position =5 ~J So, maximum KE =9 ~J -5 ~J =4 ~J Now, in SHM Maximum (at mean) KE = Maximum PE (at extremes) aligned & 1 2 k a^2=4 ~J & k= 8 a^2 = 8 10⁻⁴ =8 10^4 ~J / m ^2 & aligned Now, time period T=2 m k =2 2 8 10^4 T= 100 ~s