A system is oscillating with undamped simple harmonic motion. Then the
This question has multiple correct options

A. average total energy per cycle of the motion is its maximum kinetic energy
B. average total energy per cycle of the motion is \[\dfrac{1}{{\sqrt 2 }}\] times its maximum kinetic energy
C. root mean square velocity is \[\dfrac{1}{{\sqrt 2 }}\] times its maximum velocity
D. mean velocity is \[\dfrac{1}{2}\] of maximum velocity

Hint: Average total energy of undamped simple harmonic motion of a system is equal to the summation of its kinetic energy and potential energy. We will be using the expression of angular velocity of this system which gives the relationship between stiffness and mass of the system.


Complete step by step solution:
Let us consider the displacement of the oscillating system with undamped simple harmonic motion is given by \[y = A\sin \left( {\omega t + x} \right)\].

Writing the equation for velocity of the given system.
\[v = A\omega \cos \left( {\omega t + x} \right)\]

Kinetic energy of the given system can be expressed as:
\[K.E. = \dfrac{1}{2}m{v^2}\]……(1)
Here m is the mass of the oscillating system.

Substitute \[A\omega \cos \left( {\omega t + x} \right)\] for \[v\] in equation (1).
\[\begin{array}{c}
K.E. = \dfrac{1}{2}m{\left[ {A\omega \cos \left( {\omega t + x} \right)} \right]^2}\\
 = \dfrac{1}{2}m{A^2}{\omega ^2}{\cos ^2}\left( {\omega t + x} \right)
\end{array}\]

Potential energy of the given system can be expressed as:
\[P.E. = \dfrac{1}{2}k{y^2}\]……(2)
Here k is the stiffness of the oscillating system.

We know that the angular velocity of an oscillating undamped simple harmonic motion is given as below:
\[\begin{array}{l}
\omega = \sqrt {\dfrac{k}{m}} \\
k = {\omega ^2}m
\end{array}\]

Substitute \[{\omega ^2}m\] for k and \[A\sin \left( {\omega t + x} \right)\] for y in equation (2).
\[\begin{array}{c}
P.E. = \dfrac{1}{2}\left( {{\omega ^2}m} \right){\left[ {A\sin \left( {\omega t + x} \right)} \right]^2}\\
 = \dfrac{1}{2}m{A^2}{\omega ^2}{\sin ^2}\left( {\omega t + x} \right)
\end{array}\]

We know that the total energy of a system is equal to the summation of kinetic energy and potential energy of that system. It means that total energy of the given oscillating system is equal to summation of its kinetic as well as potential energy.
\[T.E. = K.E. + P.E.\]

Substitute \[\left[ {\dfrac{1}{2}m{A^2}{\omega ^2}{{\cos }^2}\left( {\omega t + x} \right)} \right]\] for \[K.E.\] and \[\left[ {\dfrac{1}{2}m{A^2}{\omega ^2}{{\sin }^2}\left( {\omega t + x} \right)} \right]\] for \[P.E.\] in the above equation.
\[\begin{array}{c}
T.E. = \left[ {\dfrac{1}{2}m{A^2}{\omega ^2}{{\cos }^2}\left( {\omega t + x} \right)} \right] + \left[ {\dfrac{1}{2}m{A^2}{\omega ^2}{{\sin }^2}\left( {\omega t + x} \right)} \right]\\
 = \dfrac{1}{2}m{A^2}{\omega ^2}\left[ {{{\cos }^2}\left( {\omega t + x} \right) + {{\sin }^2}\left( {\omega t + x} \right)} \right]\\
 = \dfrac{1}{2}m{A^2}{\omega ^2}
\end{array}\]

From above expression we can conclude that the total energy of the given oscillating system is constant.

The kinetic energy is maximum at the mean position and the value of \[\cos \left( {\omega t + x} \right)\] is equal to unity.
 \[\begin{array}{c}
K.E. = \dfrac{1}{2}m{A^2}{\omega ^2}\left( 1 \right)\\
 = \dfrac{1}{2}m{A^2}{\omega ^2}
\end{array}\]

On the basis of above expressions for total energy and maximum kinetic energy it can be said that average total energy per cycle of the motion is equal to maximum kinetic energy. Hence option (A) is correct and option (B) is incorrect.

We know that the root mean square velocity is equal to the \[\dfrac{1}{{\sqrt 2 }}\] of maximum velocity, hence option (C) is correct.

Maximum velocity can be found by substituting \[\cos \left( {\omega t + x} \right)\] equal to unity in the velocity at any instant of time.
\[\begin{array}{c}
{v_{\max }} = A\omega \left( 1 \right)\\
{v_{\max }} = A\omega
\end{array}\]

We also know that for a complete cycle of oscillation mean velocity is equal to zero and option (D) is incorrect.

Therefore, we can say that the average total energy per cycle of the motion is equal to maximum kinetic energy and root mean square velocity is \[\dfrac{1}{{\sqrt 2 }}\] times its maximum velocity. Hence, option (A) and (C) are correct.


Note: Mean velocity of an oscillating undamped simple harmonic system is zero if it is under consideration for a very large period of time but we are only considering the situation for one oscillation whose mean velocity is also zero. Also it is better to remember the basic expression of root mean square velocity for similar kinds of problems.