
Two waves are given by \[{y_1} = \cos \left( {4t - 2x} \right)\] and \[{y_2} = a\sin \left( {4t - 2x + \dfrac{\pi }{4}} \right).\] The phase difference between the two waves is:
A) $\pi /4$
B) $ - \pi /4$
C) $3\pi /4$
D) $\pi /2$
Answer
218.4k+ views
Hint: 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:
$\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$
Complete step by step answer:
We’ve been given the wave function of two waves as \[{y_1} = \cos \left( {4t - 2x} \right)\]and \[{y_2} = a\sin \left( {4t - 2x + \dfrac{\pi }{4}} \right).\]
To calculate the phase difference of these two waves, they must have the same trigonometric function. So, let us change one of the waves, ${y_1}$ in a sine function.
We know that $\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$, so we can alternatively write the wave function of ${y_1}$ as
${y_1} = \sin \left( {\dfrac{\pi }{2} + 4t - 2x} \right)$
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:
$\Delta \phi = \left( {4t - 2x + \dfrac{\pi }{4}} \right) - \left( {\dfrac{\pi }{2} + 4t - 2x} \right)$
\[ \Rightarrow \Delta \phi = - \pi /4\]
Hence the phase difference of the two waves will be $ - \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 $t = 0$ in which case ${y_2}$ will have a larger phase and hence the phase difference will be calculated as the difference of phases of ${y_2}$ and ${y_1}$.
Formula used:
In this solution, we will use the following formula:
$\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$
Complete step by step answer:
We’ve been given the wave function of two waves as \[{y_1} = \cos \left( {4t - 2x} \right)\]and \[{y_2} = a\sin \left( {4t - 2x + \dfrac{\pi }{4}} \right).\]
To calculate the phase difference of these two waves, they must have the same trigonometric function. So, let us change one of the waves, ${y_1}$ in a sine function.
We know that $\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$, so we can alternatively write the wave function of ${y_1}$ as
${y_1} = \sin \left( {\dfrac{\pi }{2} + 4t - 2x} \right)$
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:
$\Delta \phi = \left( {4t - 2x + \dfrac{\pi }{4}} \right) - \left( {\dfrac{\pi }{2} + 4t - 2x} \right)$
\[ \Rightarrow \Delta \phi = - \pi /4\]
Hence the phase difference of the two waves will be $ - \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 $t = 0$ in which case ${y_2}$ will have a larger phase and hence the phase difference will be calculated as the difference of phases of ${y_2}$ and ${y_1}$.
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

