Question

The equation of an SHM is given as $x = 5\sin \left( {4\pi t} \right)$, where symbols have their usual meanings. During a half cycle $\left( {0 \leqslant t \leqslant \dfrac{T}{2}} \right)$, average acceleration (in $m/{s^2}$) of the oscillator is ($x$ is in metre and time $t$ is in second)
(A) $180\pi $
(B) Zero
(C) $120\pi $
(D) $160\pi $

Answer 1

Hint: To solve this question we have to find out the equation of the acceleration of the oscillator from the equation of the displacement given. Then we need to take its average over the given time interval to get the final answer.

Formula used: The formula used to solve this question is given by
$v = \dfrac{{dx}}{{dt}}$
$a = \dfrac{{dv}}{{dt}}$
Here, $x$ is the displacement, $v$ is the velocity, $a$ is the acceleration, and $t$ is the time.

Complete step by step solution:
The displacement of the oscillator is given as
$x = 5\sin \left( {4\pi t} \right)$
We know that the velocity of a particle is related with its displacement by
$v = \dfrac{{dx}}{{dt}}$
So the velocity of the oscillator is
$v = \dfrac{{d\left( {5\sin \left( {4\pi t} \right)} \right)}}{{dt}}$
On differentiating we get
$v = 20\pi \cos \left( {4\pi t} \right)$
Now, the acceleration is related with the velocity by
$a = \dfrac{{dv}}{{dt}}$
Therefore the acceleration of the oscillator is
$a = \dfrac{{d\left( {20\pi \cos \left( {4\pi t} \right)} \right)}}{{dt}}$ (1)
On differentiating we get
$a = - 80{\pi ^2}\sin \left( {4\pi t} \right)$
Now, we know that the average of a function $f\left( x \right)$ over an interval of $\left[ {{x_1},{x_2}} \right]$ is taken as
$f = \dfrac{{\int\limits_{{x_1}}^{{x_2}} {f\left( x \right)} dx}}{{{x_2} - {x_1}}}$
So the average acceleration $A$ of the oscillator over the half cycle $\left( {0 \leqslant t \leqslant \dfrac{T}{2}} \right)$ is given by
$A = \dfrac{{\int\limits_0^{T/2} a dt}}{{\left( {T/2 - 0} \right)}}$
$A = \dfrac{2}{T}\int\limits_0^{T/2} a dt$
Substituting (1) we get
$A = \dfrac{2}{T}\int\limits_0^{T/2} {\left[ { - 80{\pi ^2}\sin \left( {4\pi t} \right)} \right]} dt$
$A = \dfrac{{ - 160{\pi ^2}}}{T}\left[ {\dfrac{{ - \cos \left( {4\pi t} \right)}}{{4\pi }}} \right]_0^{T/2}$
Substituting the limits we get
$A = \dfrac{{160{\pi ^2}}}{{4\pi T}}\left( {\cos \left( {\dfrac{{4\pi T}}{2}} \right) - \cos \left( 0 \right)} \right)$ (2)
Now, we know that the time period is related to the angular velocity by
$T = \dfrac{{2\pi }}{\omega }$
From the equation of the SHM, the angular velocity is $\omega = 4\pi {\text{ rad/s}}$
So, the time period is
$T = \dfrac{{2\pi }}{{4\pi }} = 0.5s$
So from (2) we have the average acceleration as
$A = \dfrac{{160{\pi ^2}}}{{4\pi \left( {0.5} \right)}}\left( {\cos \left( {\dfrac{{4\pi \left( {0.5} \right)}}{2}} \right) - \cos \left( 0 \right)} \right)$
$A = 80\pi \left( {\cos \left( \pi \right) - 1} \right)$
Finally we get
$A = - 160\pi {\text{ }}m/{s^2}$
So the magnitude of the average acceleration is equal to $160\pi {\text{ }}m/{s^2}$
Hence, the correct answer is option (D).

Note:
We should not choose the average acceleration of the oscillator directly as zero, simply because the average of a sinusoidal quantity is equal to zero. The average of a sinusoidal quantity is zero in a full cycle. But here we had calculated the average in a half cycle, so it is not equal to zero.