Question

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) \((3 \pi / 4)\)

Short answer

The initial phase is (A) \(\frac{2\pi}{4}\).

Step-by-step solution

Write down the given displacement equation

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.

Differentiate the displacement equation to find the velocity equation

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

Substitute the given initial conditions in the equations

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

Solve for A and θ

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: \[\theta = \cos^{-1}(\frac{1}{\sqrt{2}})\]

Determine the initial phase in the form requested in the answer choices

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) \(\frac{2\pi}{4}\).

Key concepts

Displacement Equation

In Simple Harmonic Motion (SHM), the displacement of a particle from its mean position is described using the displacement equation. The equation is:\[x = A \cos(\omega t + \theta)\]Where:

• \(x\) is the displacement from the equilibrium position,
• \(A\) is the amplitude, which represents the maximum displacement,
• \(\omega\) is the angular frequency,
• \(t\) is time,
• \(\theta\) is the phase angle.

The displacement equation shows how the particle's position changes over time as it oscillates.
It offers insights into the motion's periodic nature.
Understanding each component helps us predict the particle's location anytime during its motion.

Angular Frequency

The concept of angular frequency \(\omega\) is key to understanding oscillatory movements like Simple Harmonic Motion (SHM). Angular frequency reflects how quickly the particle is oscillating, independent of the starting phase or path.Angular frequency is defined as \[\omega = 2\pi f\]where \(f\) is the frequency of oscillation in cycles per second (Hz).
Unlike linear frequency that uses cycles, angular frequency considers how many radians the system moves per second.
This aspect of SHM helps to determine other vital characteristics such as velocity and kinetic energy.

Phase Angle

The phase angle \(\theta\) distinguishes how far the oscillation is shifted from a standard position. It indicates at what point in its cycle the motion begins. To find the phase angle, we use values from the initial conditions of the problem. For instance, here we had initial conditions such as position \(x = 1\) cm and velocity \(v = \pi\) cm/s at time \(t = 0\).
From these, we applied:1. \[1 = A\cos(\theta)\]2. \[\pi = -A\pi \sin(\theta)\]Using these relations, knowing that \(\cos^2(\theta) + \sin^2(\theta) = 1\), allowed us to solve for \(\theta\).
The phase angle reflects the system's initial configuration in radians or fractions of \(\pi\).

Velocity Equation

The velocity equation represents how quickly the displacement changes with time during Simple Harmonic Motion. This is found by differentiating the displacement equation:\[v = \frac{dx}{dt} = -A\omega \sin(\omega t + \theta)\]Here, \(v\) is the velocity, which is dependent on:

• Amplitude \(A\)
• Angular frequency \(\omega\)
• Time \(t\)
• Phase angle \(\theta\)

The negative sign indicates that the velocity direction is opposite to the positive displacement direction.
By knowing the initial conditions, such as initial velocity, we can unravel hidden details about the system's nature and its position through the equation.