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Question: The displacement of a particle is represented by the equation \(y={{\sin }^{3}}\left( \omega t \righ...

The displacement of a particle is represented by the equation y=sin3(ωt)y={{\sin }^{3}}\left( \omega t \right). The motion is
(A) non-periodic
(B) periodic but not simple harmonic
(C ) simple harmonic with period 2πω\dfrac{2\pi }{\omega }
(D) simple harmonic with period πω\dfrac{\pi }{\omega }

Explanation

Solution

The motion which repeats themselves are called periodic motions and if the motion changes with time it is said to be non-periodic motion. For finding this differentiate the given equation of particle two time with respect to time. Thus we can identify whether it is periodic or not. Hence we will get the solution. For a simple harmonic motion the restoring force of the object in motion is directly proportional to magnitude of its displacement. All the periodic motion is not a simple harmonic motion since there is no restoring force acting along them. And all oscillatory motions are periodic, but its reverse is not true.

Complete answer:
The displacement of the particle is hence given by the equation,
y=sin3ωty={{\sin }^{3}}\omega t
y=(3sinωt4sin3ωt)4y=\dfrac{\left( 3\sin \omega t-4\sin 3\omega t \right)}{4} …………….(1)
Rearranging equation (1) thus becomes,
4y=(3sinωt4sin3ωt)4y=\left( 3\sin \omega t-4\sin 3\omega t \right) ……………..(2)
Hence differentiating the equation (2) with respect to t we get,
4dydt=3cosωt.ω4×[3cos3ωt.ω]4\dfrac{dy}{dt}=3\cos \omega t.\omega -4\times \left[ 3\cos 3\omega t.\omega \right]
Again differentiating with respect to t we get,
4d2ydt2=3ω2sinωt+12ω2sin3ωt4\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=-3{{\omega }^{2}}\sin \omega t+12{{\omega }^{2}}\sin 3\omega t
d2ydt2=3ω2sinωt+12ω2sin3ωt4\Rightarrow \dfrac{{{d}^{2}}y}{d{{t}^{2}}}=\dfrac{-3{{\omega }^{2}}\sin \omega t+12{{\omega }^{2}}\sin 3\omega t}{4} …………..(3)
We know that,
sin3θ=3sinθ4sin3θ\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta
Substituting in equation (3) we get,
d2ydt2=3ω2sinωt+12ω2[3sinωt4sin3ωt]4\Rightarrow \dfrac{{{d}^{2}}y}{d{{t}^{2}}}=\dfrac{-3{{\omega }^{2}}\sin \omega t+12{{\omega }^{2}}\left[ 3\sin \omega t-4{{\sin }^{3}}\omega t \right]}{4}
From this we can conclude that,
d2ydt2αy\dfrac{{{d}^{2}}y}{d{{t}^{2}}}\alpha y
Hence, it is not a simple harmonic motion.
Because the sine function is a periodic function.

Therefore, option (B) is correct.

Note:
The period T is the time required for one complete cycle. Also frequency is the reciprocal of the time period. That is, frequency of a periodic motion is the number of oscillations per unit time. For a simple harmonic motion the restoring force of the object in motion is directly proportional to magnitude of its displacement. All the periodic motion is not a simple harmonic motion since there is no restoring force acting along them. And all oscillatory motions are periodic, but its reverse is not true.