The particle executing simple harmonic motion has a kinetic energy `K_(0) cos^(2) omega t`. The maximum values of the potential energy and the energy are respectively

Options

A. `(K_(0))/2 and K_(0)`
B. `K_(0) and 2K_(0)`
C. `K_(0) and K_(0)`
D. `0 and 2K_(0)`

Answer

D

Explanation

K.E =`K_(0)cos^(2)omegat`
Maximum `K.E = K_(0)`
then P.E = 0
so T.E `= K_(0)+0=K_(0`
if minimum K.E = 0
Then Maximum P.E = `K_(0)`
T.E = `K_(0)`