Question

Two identical traveling waves, moving in the same direction, are out of phase by$\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{}\mathbf{rad}$. What is the amplitude of the resultant wave in terms of the common amplitude${\mathit{y}}_{\mathbf{m}}$of the two combining waves?

Short answer

The amplitude of the resultant wave is$1.41y_m$

Step-by-step solution

The given data

1. Phase difference between two waves,$\varphi =\frac{\mathrm{\pi }}{2}$
2. Propagation direction of the two waves is the same.

Understanding the concept of superposition principle

Using the superposition principle and trigonometry, we can find the amplitude of the resultant wave. The resulting wave is an algebraic sum of two waves that are interfering with each other.

Formula:

The general expression of the wave, $y\left(x,t\right)=y_m\sin \left(kx-\omega t\pm \varphi \right)$ (i)

Here, $y$is displacement, is the amplitude of the wave,$y_m$ is the angular wave number, $\omega$is the Angular frequency of the wave,$t$ is time.

Calculation of the amplitude of the resultant wave

Here we have to use the wave equation for the 1^st wave using equation (i), we get

$y=y_m\sin \left(kx-\omega t\right)$

For wave second using equation (i), we get

$y=y_m\sin \left(kx-\omega t+\varphi \right)$

Hence, the resultant wave equation, using superposition principle is given as:

$y=y_m\sin \left(kx-\omega t\right)+y_m\sin \left(kx-\omega t+\varphi \right)\frac{\delta y}{\delta x}$

By using trigonometric relation

$$\begin{gathered} y=y_m\sin \left(kx-\omega t\right)+y_m\left[\sin \left(kx-\omega t\right)\cos \left(\varphi \right)+\cos \left(kx-\omega t\right)\sin \left(\varphi \right)\right] \\ =y_m\sin \left(kx-\omega t\right)+y_m\left[1+\cos \left(\varphi \right)\right]+y_m\cos \left(kx-\omega t\right)\sin \left(\varphi \right) \\ =2y_m\sin \left(kx-\omega t\right)\left[\frac{1+\cos \left(\varphi \right)}{2}\right]+2y_m\cos \left(kx-\omega t\right)\sin \left(\frac{\varphi }{2}\right)\cos \left(\frac{\varphi }{2}\right) \\ =2y_m\sin \left(kx-\omega t\right)\left(\frac{\varphi }{2}\right)+2y_m\cos \left(kx-\omega t\right)\sin \left(\frac{\varphi }{2}\right)\cos \left(\frac{\varphi }{2}\right) \\ =2y_m\sin \left(kx-\omega t\right)\left(\frac{\varphi }{2}\right)\sin \left(kx-\omega t+\varphi \right) \end{gathered}$$

Here, given$\varphi =\frac{\mathrm{\pi }}{2}$

$y=2y_m\cos \frac{\mathrm{\pi }}{4}\left(kx-\omega t+\frac{\mathrm{\pi }}{2}\right)$

By comparing the equation we can write the new amplitude as:

$$\begin{gathered} A=2y_m\mathrm{coscos}\frac{\mathrm{\pi }}{4} \\ =1.41y_m \end{gathered}$$

Hence, the value of the resultant amplitude is$1.41y_m$