For a particle executing simple harmonic motion, the kinetic energy K is given by, K=K0 cos 2ω t. The maximum value of potential energy is

Question

For a particle executing simple harmonic motion, the kinetic energy K is given by, K=K0 cos 2ω t. The maximum value of potential energy is (A)  K0  (B) zero (C)  K0/2  (D) not obtainable

Tardigrade Admin · Accepted Answer

K max =K0= total energy As total energy remains conserved in SHM, hence when U is maximum in SHM , K=0, ie, E is also equal to U max , ie, U max =E=K0.

Q. For a particle executing simple harmonic motion, the kinetic energy $K$ is given by, $K = K_{0} cos^{2} ω t .$ The maximum value of potential energy is

$K_{0}$

zero

$K_{0} /2$

not obtainable

$K_{m a x} = K_{0} =$ total energy
As total energy remains conserved in SHM, hence when $U$ is maximum in $S H M, K = 0$ ,
ie, $E$ is also equal to
$U_{m a x}$ , ie, $U_{m a x} = E = K_{0}$ .

Q. For a particle executing simple harmonic motion, the kinetic energy K is given by, K=K0​cos2ωt. The maximum value of potential energy is

K0​