Question

What is the maximum acceleration of the particle doing the SHM?
$cm$$y = 2\sin \left[ {\dfrac{{\pi t}}{2} + \phi } \right]$, where $y$ is in ${\text{cm}}$
A. $\dfrac{\pi }{2}cm/{s^2} $
B. $\dfrac{{{\pi ^2}}}{2}cm/{s^2} $
C. $\dfrac{\pi }{4}cm/{s^2}$
D. $\dfrac{{{\pi ^2}}}{4}cm/{s^2} $

Answer 1

Hint: Acceleration of a particle executing simple harmonic motion can be given to be equal to the second order derivative of the displacement of the particle. We know that displacement of a particle executing SHM is given in terms of its amplitude and sine of its phase.

Complete step-by-step answer:
Let us first of all look into the general equation of a simple harmonic motion (SHM). If $x$is the displacement of the oscillator undergoing a simple harmonic motion at a particular instant of time $t$ then the general equation representing an SHM is,
$x = A\sin (\omega t + \phi )$
where, $A$ is the amplitude of oscillation, $\omega $ is the angular frequency and $\phi $ is the phase difference.
Now, let us compare the given equation in the question and the general equation of a simple harmonic motion, then we get,
Amplitude $A = 2cm$
Angular frequency $\omega = \dfrac{\pi }{2}{\text{rad/s}}$
It is not necessary that there is always a phase difference in a simple harmonic motion, it may happen sometimes that a phase difference may be absent.
The acceleration of the particle is equal to second order derivative of displacement with respect to time. Doing so, we get
$
  a = \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{{{d^2}}}{{d{t^2}}}\left\{ {A\sin \left( {\omega t + \phi } \right)} \right\} \\
   = \dfrac{d}{{dt}}\left\{ {A\omega \cos \left( {\omega t + \phi } \right)} \right\} \\
   = - A{\omega ^2}\sin \left( {\omega t + \phi } \right) \\
$
Now the maximum acceleration of the particle is equal to the amplitude of the particle which is given as
$a = {\omega ^2}A$
Now, let us substitute the values for the amplitude and the angular acceleration as obtained by comparing the two equations, we get,
$
  a = {\left( {\dfrac{\pi }{2}} \right)^2} \times 2 \\
   \Rightarrow a = \dfrac{{{\pi ^2}}}{4} \times 2 \\
   \Rightarrow a = \dfrac{{{\pi ^2}}}{2}cm/s \\
$
Therefore from the above calculations, we can find that the maximum acceleration of the motion is given by option B.

Note: Students must be careful while comparing the equation given in the question to that of the general equation of SHM. Sometimes, instead of the value of angular frequency, the value of time may be given, in that case, we need to compare the coefficients of the angular frequency and not time.