Question:
A wave represented by the equation
\[
y=a\cos(kx-\omega t)
\]
is superposed with another to form a stationary wave such that point \(x=0\) is a node. What is the equation for the other wave?
A wave represented by the equation
\[
y=a\cos(kx-\omega t)
\]
is superposed with another to form a stationary wave such that point \(x=0\) is a node. What is the equation for the other wave?
Show Hint
If a stationary wave has a node at \(x=0\), its form is
\[
y=2a\sin kx \sin\omega t.
\]
Work backwards to identify the two travelling waves.
Updated On: Jun 16, 2026
- \[ y=-a\sin(kx+\omega t) \]
- \[ y=-a\cos(kx-\omega t) \]
- \[ y=a\cos(kx+\omega t) \]
- \[ y=-a\cos(kx+\omega t) \]
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
A stationary wave is formed by superposition of two waves of equal amplitude and frequency travelling in opposite directions.
Given wave:
\[
y_1=a\cos(kx-\omega t).
\]
The second wave must be of the form
\[
y_2=A\cos(kx+\omega t).
\]
Step 1: Use the node condition at \(x=0\). For option (D), \[ y_2=-a\cos(kx+\omega t). \] Thus \[ y=y_1+y_2 \] \[ =a\cos(kx-\omega t)-a\cos(kx+\omega t). \] Using \[ \cos A-\cos B = -2\sin\frac{A+B}{2} \sin\frac{A-B}{2}, \] \[ y = 2a\sin kx \sin\omega t. \]
Step 2: Check the displacement at \(x=0\). \[ y(0,t) = 2a\sin0\sin\omega t = 0. \] Hence \(x=0\) is always a node. \[\begin{aligned} \boxed{y=-a\cos(kx+\omega t)} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.
Step 1: Use the node condition at \(x=0\). For option (D), \[ y_2=-a\cos(kx+\omega t). \] Thus \[ y=y_1+y_2 \] \[ =a\cos(kx-\omega t)-a\cos(kx+\omega t). \] Using \[ \cos A-\cos B = -2\sin\frac{A+B}{2} \sin\frac{A-B}{2}, \] \[ y = 2a\sin kx \sin\omega t. \]
Step 2: Check the displacement at \(x=0\). \[ y(0,t) = 2a\sin0\sin\omega t = 0. \] Hence \(x=0\) is always a node. \[\begin{aligned} \boxed{y=-a\cos(kx+\omega t)} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.
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