Question

How will standing waves result from the superposition of two waves?

Answer 1

Standing wave:

1. A standing wave is formed by the superposition of two waves traveling in the opposite direction.
2. The amplitude and the face difference of the waves are the same.
3. A standing wave is represented by the form, $y(x,t)=2A\sin \left(kx\right)\sin \left(\omega t\right)$, where $A$ is the amplitude, $\omega$ is the angular frequency, $k$ is the wavenumber of the wave.

Formula used:

The resultant wave of two waves is given by the formula $y\left(x,t\right)=y_1\left(x,t\right)+y_2\left(x,t\right)$.

Formation of the standing wave:

Suppose two standing waves $y_1(x,t)=A\cos (kx-\omega t)$ and $y_2(x,t)=-A\cos (kx+\omega t)$ are directed opposite each other. Using the formula, the resultant wave is given by

$\begin{array}{rcl}y(x,t)& =& \left[A\cos \left(kx-\omega t\right)\right]+\left[-A\cos \left(kx+\omega t\right)\right]\\ & =& A\left[\cos \left(kx-\omega t\right)-\cos \left(kx+\omega t\right)\right]\\ & =& 2A\sin \frac{\left(kx-\omega t\right)+\left(kx+\omega t\right)}{2}·\sin \frac{\left(kx+\omega t\right)-\left(kx-\omega t\right)}{2}\\ & =& 2A\sin \frac{2k\omega }{2}·\sin \frac{2kx}{2}\\ & =& 2A\sin \left(kx\right)\sin \left(\omega t\right)\\ & =& A\text{'}\sin \left(\omega t\right)\end{array}$ { $\cos A-\cos B=2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right)$}

Where, $A\text{'}=2A\sin \left(kx\right)$ is the amplitude of the resultant wave.

Hence, the equation $\begin{array}{rcl}y(x,t)& =& A\text{'}\sin \left(\omega t\right)\end{array}$ is known as the equation of standing wave and represent a simple harmonic motion of the same frequency of the individual wave.