Solveeit Logo

Question

Question: Two waves are given by \[{y_1} = \cos \left( {4t - 2x} \right)\] and \[{y_2} = a\sin \left( {4t - 2x...

Two waves are given by y1=cos(4t2x){y_1} = \cos \left( {4t - 2x} \right) and y2=asin(4t2x+π4).{y_2} = a\sin \left( {4t - 2x + \dfrac{\pi }{4}} \right). The phase difference between the two waves is:
A) π/4\pi /4
B) π/4 - \pi /4
C) 3π/43\pi /4
D) π/2\pi /2

Explanation

Solution

In this solution, we will be using the general form of the wave equation and the definition of phase. To calculate the phase difference between two waves, they must be represented using the same trigonometric functions. The terms like angular frequency, wave number will be compared to find the total phase difference.

Formula used:
In this solution, we will use the following formula:
cosθ=sin(π2+θ)\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)

Complete step by step answer:
We’ve been given the wave function of two waves as y1=cos(4t2x){y_1} = \cos \left( {4t - 2x} \right)and y2=asin(4t2x+π4).{y_2} = a\sin \left( {4t - 2x + \dfrac{\pi }{4}} \right).
To calculate the phase difference of these two waves, they must have the same trigonometric function. So, let us change one of the waves, y1{y_1} in a sine function.
We know that cosθ=sin(π2+θ)\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right), so we can alternatively write the wave function of y1{y_1} as
y1=sin(π2+4t2x){y_1} = \sin \left( {\dfrac{\pi }{2} + 4t - 2x} \right)
We can now calculate the phase difference of the two waves using the difference of the variables of the trigonometric functions as the difference of the phase of the second wave and the first wave as:
Δϕ=(4t2x+π4)(π2+4t2x)\Delta \phi = \left( {4t - 2x + \dfrac{\pi }{4}} \right) - \left( {\dfrac{\pi }{2} + 4t - 2x} \right)
Δϕ=π/4\Rightarrow \Delta \phi = - \pi /4

Hence the phase difference of the two waves will be π4 - \dfrac{\pi }{4} which corresponds to option (B).

Note: We must convert the wave function of both the waves before comparing their individual phase otherwise we will get the wrong answer. Usually, the magnitude of the phase difference is considered when calculating the phase difference of the two waves. By convention, however, the phase difference is calculated taking into account which wave is at a larger displacement. To decide this, we can place the value of t=0t = 0 in which case y2{y_2} will have a larger phase and hence the phase difference will be calculated as the difference of phases of y2{y_2} and y1{y_1}.