Question

The average acceleration in one time period in a simple harmonic motion is:
A. \[A{\omega ^2}\]
B. \[A{\omega ^2}/2\]
C. \[A{\omega ^2}/\sqrt 2 \]
D. Zero

Answer 1

Hint:Simple harmonic motion or SHM is the to and fro motion of an object about the mean position. This mean position is known as the equilibrium position. One complete cycle in an SHM is when the particle returns to the mean position from one or more positions of unstable equilibrium. In this case, the net displacement becomes zero.

Formula used:
Average velocity, \[{v_{avg}}\]= Total displacement / Total time
Average acceleration, a = \[\dfrac{\text{Average velocity}}{\text{Time period}}\]=\[\dfrac{{{v_{avg}}}}{T}\]
Time period = \[T = \dfrac{{2\pi }}{\omega }\]
where \[{v_{avg}}\] = average velocity and \[\omega \]= angular velocity,

Complete step by step solution:
Given: The acceleration is to be calculated in one time period.

The time period in a simple harmonic motion is defined as the shortest time taken by an object to complete oscillation or the minimum time after which the object repeats its motion. An oscillation is the to and fro motion of an object about a mean position or the equilibrium position. An object makes one complete oscillation when it returns to its initial equilibrium position from the same direction.

According to the question, in one time period, the net displacement will be zero as the initial and final position of the object is the same. Hence average velocity and thus average acceleration will also be zero.

Hence option (D) is the correct answer.

Note: Not all oscillations are simple harmonic. The motion is simple harmonic if the mean position of the oscillation is a stable equilibrium and the restoring force is directly proportional to the displacement i.e., of the form F = -kx.