Question

The average energy in one time period in simple harmonic motion is:
(A) $\dfrac{1}{2}m{\omega ^2}{A^2}$
(B) $\dfrac{1}{4}m{\omega ^2}{A^2}$
(C) $m{\omega ^2}{A^2}$
(D) Zero

Answer 1

Hint: To answer this question we have to first begin the answer with writing the formula for finding the simple harmonic energy. The simple harmonic energy is defined as the summation of the kinetic energy and that of potential energy. From the formula we will get the expression solving which we can obtain the answer to the required question.

Complete step by step answer:
We should know that the average energy means the total energy of the simple harmonic motion in one time period.
As we know that energy is the scalar quantity and not the vector quantity, thus it will only depend on the magnitude and not on the direction.
Hence it cannot be zero as in this situation of the velocity, displacement and the others. It should be having some numerical terms as well.
Hence we can say that the total energy of the simple harmonic motion is the summation of kinetic energy and potential energy.
In formula we can write this as:
Simple Harmonic Motion = Kinetic Energy + Potential Energy
Now we have to put the values from the question into the above expression. Hence the expression becomes:
$
  \dfrac{1}{2}k({A^2} - {x^2}) + \dfrac{1}{2}k{x^2} \\
   = \dfrac{1}{2}k{A^2} - \dfrac{1}{2}k{x^2} + \dfrac{1}{2}k{x^2} \\
   = \dfrac{1}{2}k{A^2} \\
   = \dfrac{1}{2}m{\omega ^2}{A^2} \\
 $
So we can say that the average energy in one time period in simple harmonic motion is $\dfrac{1}{2}m{\omega ^2}{A^2}$.

Hence the correct answer is option A.

Note: Kinetic energy is defined as the energy that a body possesses due to the motion of the body. It can be expressed as the work that is needed to accelerate a body for a given mass from the rest to where it started velocity.
On the other hand potential energy is defined as the energy that is stored within a body, due to the position of the object, the arrangement or we can say the state of the body.