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Assertion (A): The phase difference between displacement and velocity in SHM is \[90^\circ \]
Reason (R): The displacement is represented by \[y = A\sin \omega t\] and velocity by \[V = A\omega \cos \omega t\] .
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true and R is false
(D) A is false and R is true

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Last updated date: 16th Jun 2024
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Answer
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Hint: First of all, we will differentiate the displacement with respect to the time to find the velocity. Then we will rearrange the trigonometric ratio in it to find the phase difference.

Complete step by step answer:
In the given question, we are supplied the following data:
The phase difference between the displacement and the velocity in simple harmonic motion is \[90^\circ \] .
Again, there are two equations given, one for the displacement and the other for the velocity.
The equation for the displacement is \[y = A\sin \omega t\] .
The equation for the velocity is \[V = A\omega \cos \omega t\] .
We know, in a simple harmonic motion, the velocity is maximum at the mean position, and at that point the displacement is zero.
However, at the extreme position, the velocity becomes zero while the displacement becomes maximum.
Since, the displacement is represented by the equation \[y = A\sin \omega t\] , we can find out the velocity by differentiating with respect to time. Here we go:
\[
\Rightarrow v = \dfrac{{dy}}{{dt}} \\
\Rightarrow v = \dfrac{d}{{dt}}\left( {A\sin \omega t} \right) \\
\Rightarrow v = A\omega \cos \omega t \\
\]
However, we can write:
\[\Rightarrow\cos \omega t = \sin \left( {\omega t + \dfrac{\pi }{2}} \right)\]
Both the values are the same according to trigonometry.
Now, we can write the equation of velocity as follows:
\[\Rightarrow v = A\omega \cos \omega t \\
\Rightarrow v = A\omega \sin \left( {\omega t + \dfrac{\pi }{2}} \right) \\
\]
From the above manipulation, we can say that the phase difference between the displacement and the velocity is \[\dfrac{\pi }{2}\] .
We can alternatively say that the phase difference between the displacement and the velocity is \[90^\circ \] .
Now, we can come to the conclusion that the phase difference between displacement and velocity in SHM is \[90^\circ \] where the displacement is represented by \[y = A\sin \omega t\] and velocity by \[V = A\omega \cos \omega t\] .
Both A and R are true and R is the correct explanation of A.

The correct option is A.

Note: It is important to note that in the differentiation part, many students tend to make mistakes. Remember that differentiation of sine is positive cosine and differentiation of cosine is negative sine. Phase difference tells us the separation between the wave points on the same waveform.