Question

The displacement x of a particle in motion is given in terms of time by $x\cos \omega t$
A) The particle executes SHM
B) The particle executes oscillatory motion which is not SHM
C) The motion of the particle is neither oscillator; nor simple harmonic
D) The particle is not acted upon by a force when it is at $x = 4$

Answer 1

The particle displacement of motion can be known when the equation of its motion given in terms of time and by deriving the equation of motion executed by the particle can be known. Thus, the derived equation will describe the motion of the particle.

Formula used:
Acceleration in simple Harmonic motion (SHM)
$a = - {\omega ^2}x$

Complete step by step solution:
The displacement $x$ of a particle in motion is given in terms of time is, $x\left( {x - 4} \right)\; = \;1 - 5\cos \omega t$
${x^2} - 4x = 1 - 5\cos \omega t$
Add 4 on both sides
${x^2} - 4x + 4 = 4 + 1 - 5\cos \omega t$
Simplify the equation
${\left( {x - 2} \right)^2}\; = 5\left( {1 - \cos \omega t} \right)...........\left( 1 \right)$
Differentiate the equation (1)
First derivative:
$2\left( {x - 2} \right)\dfrac{{d\left( {x - 2} \right)}}{{dt}} = 5\omega \sin \omega t\;...........\left( 2 \right)$
$\dfrac{{d\left( {x - 2} \right)}}{{dt}} = \dfrac{{5\omega \sin \omega t}}{{2\left( {x - 2} \right)}}\;..........\left( 3 \right)$
Differentiate the equation (2)
Second Derivative:

$$2\left( {x - 2} \right)\;\dfrac{{d\left( {x - 2} \right)}}{{dt}}\; = 5\omega \sin \omega t \\\ 2\left[ {\left( {\dfrac{{d{{\left( {x - 2} \right)}^2}}}{{dt}}} \right) + \;\dfrac{{\left( {x - 2} \right)\;{d^2}\left( {x - 2} \right)}}{{d{t^2}}}\;} \right] = \;5\;{\omega ^2}\cos \omega t\;...........\left( 4 \right)\; \\\$$

Substitute the value of equation (3) in the equation (4)
$2\;\left[ {{{\left( {\dfrac{{5\;\omega \sin \omega t}}{{2\left( {x - 2} \right)}}} \right)}^{^2}} + \;\dfrac{{\left( {x - 2} \right)\;{d^2}\;\left( {x - 2} \right)}}{{d{t^2}}}} \right] = 5\;{\omega ^2}\cos \omega t\;$
Displace the equation
$\left( {x - 2} \right)\dfrac{{{d^2}\left( {x - 2} \right)}}{{d{t^2}}}\; = \;\dfrac{{5\;{\omega ^2}\cos \omega t}}{2} - \dfrac{{{5^2}{\omega ^2}\left( {1 - {{\cos }^2}\omega t} \right)}}{{4{{\left( {x - 2} \right)}^2}}}$
From the equation (1)

$${\left( {x - 2} \right)^{^2}} = 5\left( {1 - \cos \omega t} \right) \\\ \- \dfrac{{{{\left( {x - 2} \right)}^2}}}{5} + 1 = \cos \omega t \\\$$

Therefore,
$\dfrac{d^2(x-2)}{dt^2}=\dfrac{-{\omega}^2(x-2)}{4}$
$\therefore \dfrac{{{d^2}x}}{{d{t^2}}}$ is the acceleration $a$
Thus,

$$a = - {\omega _0}^2x \\\ \\\$$

So, the time period, $T = \dfrac{{2\pi }}{\omega }$
The derived acceleration is in the form of acceleration of simple Harmonic motion.

Simple Harmonic Motion:
It is defined as the periodic motion of a point along a straight line because its acceleration is always towards the fixed point in that line and it is proportional to its distance from that point.

$\therefore$ The displacement $x$ of a particle in motion is given in terms of time by $x\left( {x - 4} \right) = 1 - 5\cos \omega t$ at which the particle executes SHM. Hence, option (A) is correct.

Note:
The velocity constantly changes in the simple harmonic motion.so when the velocity is zero, then the displacement is maximum when the velocity is maximum then the displacement becomes zero. The velocity of the SHM is given by $v = A\omega \sin \omega t$, $A\omega$ is the maximum speed.