Question

A particle executes two types of SHM- ${x_1} = {A_1}\sin \omega t$ and ${x_2} = {A_2}\sin \left( {\omega t + \dfrac{\Pi }{3}} \right)$ then,
(A) Find the displacement at time $t = 0$
(B) Find the maximum acceleration of the particle.
(C) Find the maximum speed of the particle.

Answer 1

Hint
One of the basic concepts required to solve this problem is the principle of superposition. Once we use this principle to calculate the resultant displacement of the particle, we can easily calculate the velocity and acceleration through differentiation.

Complete step by step answer
Vibrations which are under the action of a restoring force directly proportional to the displacement is the most fundamental type of motion in the universe and is called a simple harmonic motion. A particle executing SHM is called a Linear Harmonic Oscillator (LHO).
In this question we are told that a particle executes two different types of SHM. In such a case, the resultant motion can be obtained by using the principle of superposition. This principle states that the resultant displacement due to a number of sources is given by the algebraic sum of the displacements caused by the individual sources.
So, by using the principle of superposition we get,
$\Rightarrow x = {x_1} + {x_2}$
$\Rightarrow x = {A_1}\sin \omega t + {A_2}\sin \left( {\omega t + \dfrac{\Pi }{3}} \right)$
$\Rightarrow x = {A_1}\sin \omega t + {A_2}\sin \omega t\cos \dfrac{\Pi }{3} + {A_2}\cos \omega t\sin \dfrac{\Pi }{3}$
$\Rightarrow x = {A_1}\sin \omega t + \dfrac{1}{2}{A_2}\sin \omega t + \dfrac{{\sqrt 3 }}{2}{A_2}\cos \omega t$
$\Rightarrow x = \sin \omega t\left( {{A_1} + \dfrac{{{A_2}}}{2}} \right) + \dfrac{{\sqrt 3 }}{2}{A_2}\cos \omega t$
At $t = 0,$
$\Rightarrow x = \dfrac{{\sqrt 3 }}{2}{A_2}$
Velocity can be calculated by differentiating displacement with respect to time,
$\Rightarrow v = \dfrac{{dx}}{{dt}}$
$\Rightarrow v = \omega \left( {{A_1} + \dfrac{{{A_2}}}{2}} \right)\cos \omega t - \omega \dfrac{{\sqrt 3 }}{2}{A_2}\sin \omega t$
At $t = 0,$
$\Rightarrow v = \omega \left( {{A_1} + \dfrac{{{A_2}}}{2}} \right)$
Acceleration can be calculated by differentiating velocity with respect to time,
$\Rightarrow a = \dfrac{{dv}}{{dt}}$
$\Rightarrow a = - {\omega ^2}\left( {{A_1} + \dfrac{{{A_2}}}{2}} \right)\sin \omega t - {\omega ^2}\dfrac{{\sqrt 3 }}{2}{A_2}\cos \omega t$
At $t = 0,$
$\Rightarrow a = - {\omega ^2}\dfrac{{\sqrt 3 }}{2}{A_2}$
Taking the absolute value of acceleration we get,
$\Rightarrow a = {\omega ^2}\dfrac{{\sqrt 3 }}{2}{A_2}$
Maximum acceleration and velocity occurs at positions of maximum displacement.

Note
This kind of phenomenon is often observed in physical situations where a system is subjected to two or more simple harmonic oscillations. Most popularly, our ear membrane is often subjected to a number of vibrations all at the same time. In such a case we use the principle of superposition to calculate the resultant displacement.