Question

A wave represented by the equation $y=a\cos \left(kx-\omega t\right)$ is superposed with another wave to form a stationary wave such that point $x=0$ is a node. The equation for the other wave is

• A. $y=a\sin \left(kx+\omega t\right)$
• B. $y=-a\cos \left(kx-\omega t\right)$
• C. $y=-a\cos \left(kx+\omega t\right)$
• D. $y=-a\sin \left(kx-\omega t\right)$

Answer 1

The correct option is C

$y=-a\cos \left(kx+\omega t\right)$

Step 1. Given,

The equation of wave $y=a\cos \left(kx-\omega t\right)$

Step 2. Find the node at $x=0$

Let us consider the equation of wave,

$y=a\cos \left(kx-\omega t\right)\dots \left(1\right)$

And the other equation of wave,

$y^{\text{'}}=-a\cos \left(kx+\omega t\right)\cdots \left(2\right)$

By adding equation $\left(1\right)$ and equation $\left(2\right)$ we have,

$y+y^{\text{'}}=a\cos \left(kx-\omega t\right)-a\cos \left(kx+\omega t\right)$

Using Trignometry Formula- $\left[\cos \left(a\right)-\cos \left(b\right)=-2\sin \left(\frac{a+b}{2}\right)\sin \left(\frac{a-b}{2}\right)\right]$

$$\begin{gathered} \Rightarrow y+y^{\text{'}}=2a\sin \left(kx\right)\sin \left(\omega t\right) \\ Atx=0,\sin \left(kx\right)=0 \\ \Rightarrow y=0 \end{gathered}$$

Therefore, $x=0$ is a node.

Since we got $x=0$ node for the equation $y^{\text{'}}=-a\cos \left(kx+\omega t\right)$,

Hence, $y^{\text{'}}=-a\cos \left(kx+\omega t\right)$ is the equation of other wave.