Question

The equation of motion of a particle executing S.H.M. where letters have the usual meaning is:
A) $\dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{k}{m}x$
B) $\dfrac{{{d^2}x}}{{d{t^2}}} = + {\omega ^2}x$
C) $\dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$
D) $\dfrac{{{d^2}x}}{{d{t^2}}} = - kmx$

Answer 1

Hint:A particle executing a periodic motion where the displacement of the particle has a sinusoidal nature is said to be executing a simple harmonic motion (SHM). The equation of motion can be obtained by taking the second derivative of the displacement equation of the particle.

Complete step by step answer.
Step 1: List the displacement equation of the particle which executes simple harmonic motion (SHM).
The displacement of the particle undergoing a simple harmonic motion is a function of time and is sinusoidal. Let $x$ be the displacement of the particle in time $t$.
Then we can represent the displacement of the particle as $x\left( t \right) = A\sin \omega t$ ------- (1) where $A$ is the amplitude of the motion and $\omega $ is the angular velocity of the particle.
Step 2: Take the second derivative of equation (1) to obtain the equation of motion of the given particle.
Equation (1) is given as $x\left( t \right) = A\sin \omega t$ .
Taking the derivative of equation (1) we get, $\dfrac{{dx}}{{dt}} = \dfrac{{d\left( {A\sin \omega t} \right)}}{{dt}} = A\omega \cos \omega t$
Thus the first derivative of equation (1) is $\dfrac{{dx}}{{dt}} = A\omega \cos \omega t$ ---------- (2).
Now taking the derivative of equation (2) gives us the second derivative of equation (1).
i.e., $\dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{d}{{dt}}\left( {\dfrac{{dx}}{{dt}}} \right)$
This then becomes $\dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{{d\left( {A\omega \cos \omega t} \right)}}{{dt}} = - {\omega ^2}A\sin \omega t$ ------- (3)
On comparing equations (1) and (3) we get, $\dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$
Thus the equation of motion of the given particle is $\dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$.
Since the angular velocity is given by ${\omega ^2} = \dfrac{k}{m}$, the equation of motion can also be represented as $\dfrac{{{d^2}x}}{{d{t^2}}} = - \dfrac{k}{m}x$ .

So the correct option is C.

Note:The derivative of $\sin \omega t$ is $\dfrac{{d\left( {\sin \omega t} \right)}}{{dt}} = \cos \omega t \times \omega $ in equation (2) and the derivative of $\cos \omega t$ is $\dfrac{{d\left( {\cos \omega t} \right)}}{{dt}} = - \sin \omega t \times \omega $ in equation (3). The first derivative of equation (1) corresponds to the velocity of the particle as velocity is the rate of change of displacement i.e., $v = \dfrac{{dx}}{{dt}} = A\omega \cos \omega t$. The derivative of equation (2) then corresponds to the acceleration of the particle as acceleration is the rate of change of velocity i.e., $a = \dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$ .