Question

he equation of a simple harmonic wave is given by $y = 6\sin \left[ {2\pi \left( {2t - 0.1x} \right)} \right]$, where x and y are in millimetres and t is in seconds. The phase difference between the two particles 2 millimetres apart at any instant is
A. ${18^0}$
B. ${36^0}$
C. ${54^0}$
D. ${72^0}$

Answer 1

In this question, we need to determine the phase difference between the two particles, which are 2 millimetres apart from each other at any instant of its motion. For this, we will follow the standard equation of motion of a simple harmonic motion and compare it with the given equation to evaluate the phase difference.

Complete step by step answer: The given simple harmonic equation can be re-written as:
$y = 6\sin \left[ {2\pi \left( {2t - 0.1x} \right)} \right] \\\ = 6\sin \left( {4\pi t - 0.2\pi x} \right) \\\$
The phase displacement is the net displacement of the simple harmonic wave from its regular (or, usual) motion. Mathematically, $y = A\sin \left( {\omega t + \phi } \right)$ is the standard equation of motion of a simple harmonic wave such that ‘A’ is the amplitude or the maximum displacement of the wave on either direction of the center, $\omega$ is the angular velocity of the particle moving in a simple harmonic motion and $\phi$ is the phase difference of the particle moving in a simple harmonic motion at any two instances.
So, from the above equation, we can say that the phase displacement of the simple harmonic wave is given as $\phi = | - 0.2\pi x|$.
Let ${x_1}$ and ${x_2}$ be the phase displacement of the particle at two instances of time from the center of motion. So,
${\phi _1} = \left| { - 0.2\pi {x_1}} \right|$ and ${\phi _2} = \left| { - 0.2\pi {x_2}} \right|$.
According to the question, the particles are always 2 millimetres apart at any instant of motion.
So,${x_1} - {x_2} = 2mm$.
Substituting the values of ${\phi _1}$ and ${\phi _2}$in the equation $\vartriangle \phi = {\phi _1} - {\phi _2}$ to establish a relation between the phase difference and the position difference.

$$\vartriangle \phi = \left| {{\phi _1} - {\phi _2}} \right| \\\ = \left| { - 0.2\pi {x_1}} \right| - \left| { - 0.2\pi {x_2}} \right| \\\ = 0.2\pi \left| {{x_1} - {x_2}} \right| \\\ = 0.2 \times {180^0} \times 2 \\\ = 0.4 \times {180^0} \\\ = {72^0} \\\$$

Hence, the phase difference between the two particles 2 millimetres apart at any instant is ${72^0}$.
Option D is correct.

Note: Here, we have used the modulus for the difference in the phases as the particles are in motion between the endpoints with respect to the centre point of its motion. Moreover, students must be aware of the terms phase difference and position difference in order to solve these types of questions.