The displacement for a particle performing S.H.M. is given by \(\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\theta)\). If the initial position of the particle is \(1 \mathrm{~cm}\) and its initial velocity is $\pi \mathrm{cms}^{-1}$, then what will be its initial phase ? The angular frequency of the particle is \(\pi \mathrm{s}^{-1}\). (A) \((2 \pi / 4)\) (B) \((7 \pi / 4)\) (C) \((5 \pi / 4)\) (D) \((3 \pi / 4)\)

The displacement of the particle is given by: \[x = A \cos(\omega t + \theta)\] where x is the displacement, A is the amplitude, ω is the angular frequency, t is the time, and 𝜃 is the phase angle. We have the initial position x = 1 cm and the initial velocity v = π cm/s at t = 0s, and the angular frequency ω = π s⁻¹.

To find the equation for the velocity of the particle, differentiate the displacement equation x with respect to time, t: \[v = \frac{dx}{dt} = -A\omega \sin(\omega t + \theta)\]

At t = 0s, we know x = 1 cm, and v = π cm/s: For displacement, x: \[1 = A\cos(\theta)\] For initial velocity, v: \[\pi = -A\pi \sin(\theta)\]

From the initial velocity equation, we have: \[\pi = -A\pi \sin(\theta)\] Divide both sides by π: \[1 = -A\sin(\theta)\] Now we have a system of two equations with two unknowns, A and θ: \[1 = A\cos(\theta)\] \[1 = -A\sin(\theta)\] Square both equations: \[1 = A^2\cos^2(\theta)\] \[1 = A^2\sin^2(\theta)\] Add both equations to get: \[2 = A^2 (\cos^2(\theta) + \sin^2(\theta))\] Since \(\cos^2(\theta) + \sin^2(\theta) = 1\), then: \[2 = A^2\] Take the square root of the equation: \[A = \sqrt{2}\] Now substitute the value of A in the first equation to find θ: \[1 = \sqrt{2} \cos(\theta)\] Divide both sides by sqrt(2): \[\frac{1}{\sqrt{2}} = \cos(\theta)\] Now, find the value of θ: \[\theta = \cos^{-1}(\frac{1}{\sqrt{2}})\]

In the answer choices, the initial phase is given in the form of a fraction of π. We can now find which one matches our obtained θ value: \[\theta = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}\] Hence the initial phase is (A) \(\frac{2\pi}{4}\).