Question

Two sound waves are y = asin(wt-kx) and y= acos(wt - kx) the phase difference between the two waves is

Answer 1

let the 2 waves be given by y1 and y2

y1=asin(wt-kx)

y2=acos(wt-kx)

now y2 can be written as

y2=asin(90-(wt-kx))

phase difference=Ф1-Ф2

=90

Answer 2

The phase difference between wave 1 and wave 2 is 90 degrees.

Explanation:

Displacement of wave 1 is :

$y_1=A\sin (\omega t-kx)$

The displacement of wave 2 is :

$y_2=A\cos(\omega t-kx)$

$\omega$ is angular frequency

k is propagation constant

First wave is in the form of sine wave and wave 2 is in the form of cos wave. We know that :

$\sin(90-\theta)=\cos \theta$

We can write wave 2 as :

$A\cos(\omega t-kx)=A\sin (90-(\omega t-kx))$

It is clear that the phase difference between wave 1 and wave 2 is 90 degrees.

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