# Two waves, traveling in the same direction through the same region, have equal frequencies,...

## Question:

Two waves, traveling in the same direction through the same region, have equal frequencies, wavelengths, and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90, what is the resultant amplitude?

The classical wave equation permits the superposition of waves. Thus any number of waves can coexist in the same region of space at the same time. The resultant displacement due to more than one wave can be determined by recourse to phasor addition. A wave has an amplitude and a phase angle. So it is just like a vector with a magnitude and a direction angle. Imagine an arrow rotating with the same frequency as that of the wave. The angle made by this with the x-axis is the phase angle. The length of the arrow is the amplitude. So for adding two waves it is enough to use the parallelogram law of addiion.

If a wave with amplitude {eq}\displaystyle {a_1} {/eq} is superposed with another wave of amplitude {eq}\displaystyle {a_2} {/eq} and if the phase difference between the waves is {eq}\displaystyle {\phi} {/eq} then the resultant wave has an amplitude,

{eq}\displaystyle { a=\sqrt{a_1^2+a_2^2+2a_1a_2 \cos \phi}}---------(1) {/eq}.

Here it is given that the two waves have an amplitude of 4 mm each and a phase difference of 90 degrees. They have the same frequency. Therefore from (1), the amplitude of the resultant wave is,

{eq}\displaystyle { \sqrt{4^2+4^2+2\times 4\times 4\times \cos90}=4\sqrt{2}\ mm} {/eq}. 