Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Answer  Hint: First of all we will see the equation of displacement of a particle executing Simple Harmonic Motion. Then, on differentiating that equation with respect to time, we get the equation of velocity. On comparing the phase constant for both the terms, that is, displacement and velocity we will be able to determine the phase difference between them at any given time instant. Complete step-by-step answer: Let the displacement of the particle from the point about which it is executing simple harmonic motion be given by the term $X$ .Now, if the particle has an initial phase of $\theta $ and angular velocity $\omega $, then the displacement equation of the particle could be written as follows:$\Rightarrow X=A\sin (\omega t+\theta )$ [Let this expression be equation number (1)]Where, A is the maximum displacement about the point of execution of simple harmonic motion. This maximum displacement is termed as Amplitude.Now, we know the differential of displacement of a particle with respect to time gives us the particle’s velocity. Thus, on differentiating equation number (1) with respect to time, we get:$\begin{align} & \Rightarrow \dfrac{dX}{dt}=\dfrac{d[A\sin (\omega t+\theta )]}{dt} \\ & \Rightarrow v=A\omega \cos (\omega t+\theta ) \\ \end{align}$ Here, we can write the term obtained for velocity as:$\Rightarrow A\omega \cos (\omega t+\theta )=-A\omega \sin (\omega t+\theta +\dfrac{\pi }{2})$ [Using: $\cos \left( \dfrac{\pi }{2}+x \right)=-\sin x$ ]$\therefore v=-A\omega \sin (\omega t+\theta +\dfrac{\pi }{2})$ [Let this expression be equation number (2)]On comparing equation number (1) and (2), we get:The difference in the phase of velocity and displacement as $\dfrac{\pi }{2}$ .Therefore, for a particle executing Simple Harmonic Motion, the phase difference between velocity and displacement is equal to $\dfrac{\pi }{2}$ radian.So, the correct answer is “Option A”. Note: We see that the difference in phase of velocity and displacement is independent of any parameter. That is, at any given instant this difference remains the same. If we differentiate the velocity equation again, we will see that the phase difference between displacement and acceleration is $\pi $ radian. |
What is the phase relationship between displacement and velocity in SHM?
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