Question

The magnitude of average acceleration in half time period in a simple harmonic motion:
A. \[\dfrac{{2{\omega ^2}A}}{T}\]
B. \[\dfrac{{{A^2}}}{{2T}}\]
C. \[\dfrac{{2A}}{{\omega T}}\]
D. Zero

Answer 1

Hint: A special type of periodic motion where the restoring force of the moving object is directly proportional to its displacement magnitude and which acts towards the objects equilibrium position is called simple harmonic motion. It is the periodic motion of a point along a straight line such that its acceleration is always towards a fixed point and it is directly proportional to its distance from that point.
It is basically repetitive movement back and forth of an object through an equilibrium or central position such that maximum displacement on one side of its position is equal to maximum displacement on the other side.

Complete step by step solution:
Let the displacement function be \[x = A\sin \omega t\]
Where
\[A\]is the amplitude
\[\omega \]is the angular frequency\[ = \dfrac{{2\pi }}{T}\]
\[T\]is time period
The average acceleration in Simple harmonic motion is given as
\[{a_{av}} = \dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}A\sin \omega t\]
Hence the average acceleration for the half time period will be
\[
  {a_{av}} = \dfrac{{\int\limits_0^{\left( {\dfrac{T}{2}} \right)} {a\left( t \right)dt} }}{{\int\limits_0^{\dfrac{T}{2}} {dt} }} \\
   = \dfrac{{\left( { - {\omega ^2}A} \right)\int\limits_0^{\left( {\dfrac{T}{2}} \right)} {\sin \left( {\omega t} \right)dt} }}{{\dfrac{T}{2}}} \\
   = \dfrac{2}{T}\left( { - {\omega ^2}A} \right) \times \dfrac{1}{\omega }\left[ { - \cos \left( {\omega t} \right)} \right]_0^{\left( {\dfrac{T}{2}} \right)} \\
   = - \dfrac{{2\omega A}}{T}\left[ {\left( { - \cos \left( {\dfrac{{\omega T}}{2}} \right)} \right) + \cos 0} \right] \\
   = - \dfrac{{2\omega A}}{T}\left[ {\left( { - \cos \left( {\dfrac{{2\pi T}}{{2T}}} \right)} \right) + 1} \right]{\text{ }}\left[ {As,\omega = \dfrac{{2\pi }}{T}} \right] \\
   = - \dfrac{{2\omega A}}{T}\left[ {\left( { - \cos \pi } \right) + 1} \right] \\
   = - \dfrac{{2\omega A}}{T}\left[ {1 + 1} \right] \\
   = - \dfrac{{4\omega A}}{T} \\
   = - \dfrac{{2{\omega ^2}A}}{\pi } \\
 \]
Hence the magnitude of average acceleration in half time period is \[\dfrac{{2{\omega ^2}A}}{T}\]

Option A is correct.

Note: The time period is the time taken for an object to complete one cycle of its periodic motion, such as time taken by a pendulum to make full back and forth swing. The magnitude of the average acceleration is the half time period for equilibrium position in a simple harmonic motion.