
The displacement of a particle along the x axis is given by $x = a{\sin ^2}\omega t$. The motion of particle corresponds to
A) Simple harmonic motion of frequency $\dfrac{\omega}{\pi}$.
B) Simple harmonic motion of frequency $\dfrac{3 \omega}{2 \pi}$.
C) Non simple harmonic motion.
D) Simple harmonic motion of frequency $\dfrac{\omega}{2 \pi}$.
Answer
139.8k+ views
Hint: For a particle to perform simple harmonic motion its acceleration should be directly proportional to the displacement i.e. $a \propto - x$ and the force should act towards the object’s equilibrium. To calculate the acceleration, the displacement is double differentiated. If the acceleration of the body is directly proportional to its displacement then the body is said to be in simple harmonic motion.
Complete step by step solution:
The answer to this question is c i.e. it is a non harmonic motion. so we have to calculate the acceleration from the given expression. The given expression $x = a{\sin ^2}\omega t$ gives the relation between displacement, amplitude and the angular velocity.
It is given that $x = a{\sin ^2}\omega t$
Applying the trigonometric equation,
$x = a\left( {{{1 - \cos 2\omega t} \over 2}} \right)$ \[\left[ {\cos 2\theta = 1 - 2{{\sin }^2}\theta } \right]\]
$x = a\left( {{{1 - \cos 2\omega t} \over 2}} \right)$
After differentiating we get, $v = {{dx} \over {dt}} = a\omega \sin \omega t$
And $acc = {{dv} \over {dt}} = 2a{\omega ^2}\cos \omega t$
Since, acceleration is not directly proportional to - $x$. Therefore this motion is not a simple harmonic motion. The simple wave is generally represented by sine waves. Therefore, its expression is a function of sine and angular frequency. Here it represents the time period of the wave.
Note: The simple harmonic motion is a special type of periodic motion where the restoring force applied on the object is directly proportional to the magnitude of displacement and acts towards the object's equilibrium position. Also all simple harmonic motion are periodic motion but all periodic motion are not simple harmonic motion.
Complete step by step solution:
The answer to this question is c i.e. it is a non harmonic motion. so we have to calculate the acceleration from the given expression. The given expression $x = a{\sin ^2}\omega t$ gives the relation between displacement, amplitude and the angular velocity.
It is given that $x = a{\sin ^2}\omega t$
Applying the trigonometric equation,
$x = a\left( {{{1 - \cos 2\omega t} \over 2}} \right)$ \[\left[ {\cos 2\theta = 1 - 2{{\sin }^2}\theta } \right]\]
$x = a\left( {{{1 - \cos 2\omega t} \over 2}} \right)$
After differentiating we get, $v = {{dx} \over {dt}} = a\omega \sin \omega t$
And $acc = {{dv} \over {dt}} = 2a{\omega ^2}\cos \omega t$
Since, acceleration is not directly proportional to - $x$. Therefore this motion is not a simple harmonic motion. The simple wave is generally represented by sine waves. Therefore, its expression is a function of sine and angular frequency. Here it represents the time period of the wave.
Note: The simple harmonic motion is a special type of periodic motion where the restoring force applied on the object is directly proportional to the magnitude of displacement and acts towards the object's equilibrium position. Also all simple harmonic motion are periodic motion but all periodic motion are not simple harmonic motion.
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