Question

# The particle executing simple harmonic motion has a kinetic energy of $${ K }_{ 0 }\cos ^{ 2 }{ \omega t }$$. The maximum values of the potential energy and the total energy are respectively

A
K0 and K0
B
0 and 2K0
C
K02 and K0
D
K0 and 2K0

Solution

## The correct option is A $${ K }_{ 0 }$$ and $${ K }_{ 0 }$$The kinetic energy of particle executing S.H.M is $$K.E={ K }_{ 0 }\cos ^{ 2 }{ \omega t }$$Maximum value $$K.E$$ is $$K.E_{ mean }={ K }_{ 0 }$$Now, the total energy of S.H.M is same as the maximum value of $$K.E$$, which is $${ K }_{ 0 }$$Similarly the maximum value of potential energy of S.H.M is same as the maximum value of $$K.E$$, which is $${ K }_{ 0 }$$ Physics

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