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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 total energy are respectively
A
$\dfrac{K_0}{2}$ and $K_0$
B
$K_0$ and $2K_0$
C
$K_0$ and $K_0$
D
0 and $2K_0$
Detailed Solution
We have $U + K = E$, where $U$ = potential energy, $K$ = kinetic energy and $E$ = total energy.
In S.H.M., when the potential energy is maximum the kinetic energy is zero, and vice-versa.
$\therefore U_{max} + 0 = E \Rightarrow U_{max} = E$
Further, K.E. $= \dfrac{1}{2}m\omega^2a^2\cos^2\omega t$
But by the question, K.E. $= K_0\cos^2\omega t$, so $K_0 = \dfrac{1}{2}m\omega^2a^2$
Hence the total energy is $E = \dfrac{1}{2}m\omega^2a^2 = K_0$
$\therefore U_{max} = K_0$ and $E = K_0$.
In S.H.M., when the potential energy is maximum the kinetic energy is zero, and vice-versa.
$\therefore U_{max} + 0 = E \Rightarrow U_{max} = E$
Further, K.E. $= \dfrac{1}{2}m\omega^2a^2\cos^2\omega t$
But by the question, K.E. $= K_0\cos^2\omega t$, so $K_0 = \dfrac{1}{2}m\omega^2a^2$
Hence the total energy is $E = \dfrac{1}{2}m\omega^2a^2 = K_0$
$\therefore U_{max} = K_0$ and $E = K_0$.