Question

When two progressive waves $y_1 = 4 \sin (2x -6t) $ and $y_2 = 3 \sin (2x -6t - \dfrac{\pi}{2}) $ are superimposed, the amplitude of the resultant wave is

Answer 1

Hint: In the given two waves we have the information about the amplitudes and phase difference between the two waves. If there was no phase difference the two waves would have been exactly similar but with different amplitudes.
Formula used:
The amplitude of resultant wave can be written as:
$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$.

Complete answer:
We consider two waves
$y_1 = a_1 \sin (q) $
$y_2 = a_2 \sin (q + \phi) $
where the q part tells us about the space propagation and time propagation of the wave.

When the two waves superpose, we get:
$Y = y_1 + y_2 = A \sin (q + \theta)$
where $\theta$ and A are constants in the new wave formed.
The amplitude of the resultant wave can be written as:
$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$
where $\phi$ is the phase difference between the two waves.

The Waves given to us in the question;
$y_1 = 4 \sin (2x -6t) $
and
$y_2 = 3 \sin (2x -6t - \dfrac{\pi}{2}) $
look quite similar to the two we considered in the amplitude expression.
If we compare the two, we get
$a_1 = 4 $,
$a_2 = 3 $ and
$\phi = - \dfrac{\pi}{2}$.
Keeping these in the expression for A, we get
$A = \sqrt{4^2 + 3^2 + 0} = 5$
Therefore, the amplitude of the resultant wave formed by superposition of two given waves is 5 units.

Additional Information:
Two light waves traveling in the same direction, with the same frequency and having fixed phase difference between them, cause interference of light. The interference of light is an important phenomenon as it rigidly establishes the wave nature of light.

Note: The sum of the expressions that we used for two waves can be used to determine the amplitude by performing some algebra. Also, the coefficient of t in the given two waves is the angular frequency of the wave and the coefficient of x in the given waves is propagation constant k.