Question

The equation of the displacement of two particles making SHM are represented by${{y}_{1}}=a\sin (\omega t+\phi )$ &${{y}_{2}}=a\cos (\omega t)$. The phase difference of the velocities of the two particles is:

& A.\,\dfrac{\pi }{2}+\phi \\\ & B.\,-\phi \\\ & C.\,\phi \\\ & D.\,\phi -\dfrac{\pi }{2} \\\ \end{aligned}$$

Answer 1

The given equations represent the equation of the displacement of two particles making SHM, but, we are asked to find the phase difference of the velocities of the two particles, so, firstly, we will differentiate the equation and will then find the phase difference.
Formula used:
$P={{P}_{0}}+\rho gh$

Complete answer:
From the given information, we have the data as follows.
The equation of the displacement of two particles making SHM are represented by${{y}_{1}}=a\sin (\omega t+\phi )$ &${{y}_{2}}=a\cos (\omega t)$.
Firstly, we will convert the given equations of displacement into the equations of velocity. In order to do so, we have to differentiate the equations.
Consider the displacement equations of two particles making SHM.
${{y}_{1}}=a\sin (\omega t+\phi )$ &${{y}_{2}}=a\cos (\omega t)$.
Now differentiate the above equations with respect to the time.

& \dfrac{d{{y}_{1}}}{dt}=a\omega \cos (\omega t+\phi ) \\\ & \therefore {{v}_{1}}=a\omega \cos (\omega t+\phi ) \\\ \end{aligned}$$ And $$\begin{aligned} & \dfrac{d{{y}_{2}}}{dt}=a\cos (\omega t) \\\ & \therefore {{v}_{2}}=-a\omega \sin (\omega t) \\\ \end{aligned}$$ As both of the equations of the velocity of two particles making SHM have different angle functions, so, we need to convert one of the equations to have the same angle function. Consider the second equation of the velocity. $${{v}_{2}}=a\omega \cos \left( \omega t-\dfrac{\pi }{2} \right)$$ Therefore, the equations of the velocity of two particles making SHM are represented as follows. $${{v}_{1}}=a\omega \cos \left( \omega t+\phi \right)$$ and $${{v}_{2}}=a\omega \cos \left( \omega t-\dfrac{\pi }{2} \right)$$ Now, we will find the phase difference of these velocity equations. $$\begin{aligned} & pd=\left( \omega t+\phi \right)-\left( \omega t-\dfrac{\pi }{2} \right) \\\ & \therefore pd=\phi +\dfrac{\pi }{2} \\\ \end{aligned}$$ $$\therefore $$ The phase difference of the velocities of the two particles is $$\phi +\dfrac{\pi }{2}$$. **Thus, option (A) is correct.** **Note:** The differentiation of the displacement gives the velocity. The trigonometric equations in terms of the angles should be the same, that is, either we have to convert the angles in the form of a sine angle or in the form of the cosine angle. While computing the phase difference, we have to consider only the angle values and not the whole wave equation.