Question

A particle is executing SHM along the x-axis given by $x=A\sin ot$. What is the magnitude of the average acceleration of the particle between $t=0$ and $t$=${\displaystyle \frac{T}{4}}$, where $T$ is the time period of oscillation.
(A) ${\displaystyle \frac{2{\omega }^{2}A}{\pi }}$
(B)${\displaystyle \frac{{\omega }^{2}A}{\pi }}$
(C)${\displaystyle \frac{4{\omega }^{2}A}{\pi }}$
(D) None of these

Answer 1

Hint: According to the question we need to first calculate the derivative of a given equation to find the velocity of the particle with respect to time. Then we need to calculate the average acceleration by putting the formula from the first law of motion. The average acceleration of a particle is equal to ${\displaystyle \frac{v-u}{t}}$

Complete answer:
Any particle executing simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the magnitude of the object ‘ s displacement and acts towards the equilibrium position .
Here, the motion of the particle is represented by
$x(t)=A\sin \omega t$
Where, $A$is Amplitude of the motion,$\omega$ is the angular frequency of the motion (‘o’ in the question) and $t$ represents time.
Now, we need to calculate the velocity of the particle by differentiating the given equation with respect to time.
$v={\displaystyle \frac{dx(t)}{dt}}=-A\omega \cos \omega t$
Here, $v$ represents velocity of the particle at any instant of time.
From the first law of motion, we know Average acceleration of any rigid body in motion is given by $a$= ${\displaystyle \frac{v-u}{t}}$
Here, $v$ is the velocity of the particle at time = ${\displaystyle \frac{T}{4}}$ and
$u$ is the velocity of the particle at time = $0$
Now, putting the respective values of $v$ and $u$ after differentiation and putting total time as ${\displaystyle \frac{T}{4}}$
    =${\displaystyle \frac{-A\omega \cos \omega t_{t={\displaystyle \frac{T}{4}}}+A\omega \cos \omega t_{t=0}}{{\displaystyle \frac{T}{4}}}}$
    =${\displaystyle \frac{A\omega }{{\displaystyle \frac{T}{4}}}}$
    =${\displaystyle \frac{A\omega }{{\displaystyle \frac{2\pi }{4\omega }}}}$
    =${\displaystyle \frac{4A\omega \cdot \omega }{2\pi }}$
    =${\displaystyle \frac{2A{\omega }^2}{\pi }}$

Note:
We should take into consideration that the average acceleration of the particle shouldn’t be taken as a double derivative of position with respect to time. Make sure to put the values of time as ${\displaystyle \frac{T}{4}}$ in $v$ and $0$in $u$after differentiation.