A linear harmonic oscillator of force constant 2×106 N/m and amplitude 0.01 m has a total mechanical energy 160 J. Its
The correct option is B,C Maximum K.E. is 100 J and maximum potential energy is 160 J
As it is given, TE=KE+PE = 160 J
160=KEmax+PEmin (at mean position)
160=0+PEmax
PEmax=160J
KEmax=12KA2=12×2×106×(0.01)2=100 J