If the equation for displacement of two particles executing S.H.M. is given by \(\mathrm{y}_{1}=2 \sin (10 \mathrm{t}+\theta)\) and $\mathrm{y}_{2}=3 \cos 10 \mathrm{t}$ respectively, then the phase difference between the velocity of two particles will be \(\ldots \ldots \ldots\) (A) \(-\theta\) (B) \(\theta\) (C) \(\theta-(\pi / 2)\) (D) \(\theta+(\pi / 2)\).

To find the velocities of the particles, we will differentiate the displacement equations with respect to time: Particle 1: The displacement equation is given by \(y_1 = 2 \sin(10t + \theta)\) Taking the derivative of \(y_1\) with respect to time \(t\) gives the velocity: \(v_1 = \frac{dy_1}{dt} = 2 \cdot 10 \cos(10t + \theta) = 20 \cos(10t + \theta)\) Particle 2: The displacement equation is given by \(y_2 = 3 \cos(10t)\) Taking the derivative of \(y_2\) with respect to time \(t\) gives the velocity: \(v_2 = \frac{dy_2}{dt} = -3 \cdot 10 \sin(10t) = -30 \sin(10t)\)

Now, we want to calculate the phase difference between these two velocities. Let's rewrite the velocities as a product of a trigonometric function and a constant to express it as a simple harmonic equation: \(v_1 = 20 \cos(10t + \theta) = 20 \cos (\omega t + \phi_1)\) with \(\omega = 10\) and \(\phi_1 = \theta\) \(v_2 = -30 \sin(10t) = -30 \cos \left(10t - \frac{\pi}{2}\right) = 30 \cos (\omega t + \phi_2)\) with \(\omega = 10\) and \(\phi_2 = -\frac{\pi}{2}\) The phase difference between the two velocities can be calculated by subtracting \(\phi_1\) from \(\phi_2\): \(\Delta \phi = \phi_2 - \phi_1 = (-\frac{\pi}{2}) - \theta\) Therefore, the phase difference between the velocity of two particles is: \(\Delta \phi = \theta - (\frac{\pi}{2})\) The correct answer is (C).