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# 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) $\pi /2$

Last updated date: 13th Jun 2024
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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:
$\cos \theta = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$

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
${y_1} = \sin \left( {\dfrac{\pi }{2} + 4t - 2x} \right)$
$\Delta \phi = \left( {4t - 2x + \dfrac{\pi }{4}} \right) - \left( {\dfrac{\pi }{2} + 4t - 2x} \right)$
$\Rightarrow \Delta \phi = - \pi /4$
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 ${y_1}$.