Animated Solution for Physics - Waves: A wave represented by the equation $y=a\cos (kx-\omega t)$ is superimposed with another wave to form a stationary wave such that point $x=0$ is a node. The equation for the other wave is

Standing waves, or stationary waves, represent one of the most visually stunning and physically profound phenomena in wave mechanics. Unlike progressive waves, which transport energy continuously through a medium, standing waves seem to "stand still" in space. They are characterized by points of absolute rest, called nodes, and points of maximum oscillation, called antinodes.

But how do these stationary patterns emerge from dynamic, moving waves? The answer lies in the principle of superposition. When two identical progressive waves traveling in opposite directions overlap, their interference creates a stationary pattern. In this article, we will dissect a classic JEE problem from 1988 that explores the mathematical conditions required to form a standing wave with a node at a specific boundary.

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We are given a progressive wave represented by the equation:

Let's analyze this equation to understand its physical behavior: 1. Amplitude: The maximum displacement of the particles is $a$. 2. Propagation Direction: The phase term is $(kx-\omega t)$. Since the spatial term $kx$ and the temporal term $\omega t$ have opposite signs, this wave is propagating in the positive $x$-direction. 3. Frequency and Wavelength: The angular frequency is $\omega$ and the wave number is $k$.

To form a standing wave, we must superimpose this wave with another wave, $y_2$, that satisfies two strict conditions: - It must have the same amplitude $a$, frequency $\omega$, and wave number $k$ to ensure perfect cancellation and reinforcement. - It must travel in the opposite direction (the negative $x$-direction).

Therefore, the general equation for the second wave must be of the form:

where the positive sign between $kx$ and $\omega t$ ensures propagation in the negative $x$-direction, and $\varphi$ is an arbitrary phase constant that we must determine.

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Node at $x=0$

A node is a point in the medium that undergoes absolutely zero displacement at all times. Mathematically, this means the net displacement $y_{\text{net}}$ at $x=0$ must be identically zero for any time $t$:

Let's substitute $x=0$ into our individual wave equations:

Since cosine is an even function, i.e., $\cos (-\theta )=\cos (\theta )$, we can simplify this to:

Now, substituting $x=0$ into our general equation for $y_2$:

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Now, we apply the superposition principle at the origin:

Rearranging this equation gives:

For this equality to hold true at every single instant of time $t$, the two cosine functions must be completely out of phase (shifted by $180^{\circ }$ or $\pi$ radians), and their amplitudes must be equal. This yields:

1. Amplitude: $A=a$ 2. Phase Constant: $\varphi =\pi$

Let's substitute these values back into our general equation for $y_2$:

Using the trigonometric identity $\cos (\theta +\pi )=-\cos (\theta )$, we can simplify this to:

This is a remarkably elegant result! It shows that the second wave must not only travel in the opposite direction but must also be inverted (shifted by $\pi$ radians) to ensure that they always cancel each other out at the origin.

Comparing this with our options, we find that it matches Option (c).

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What if $x=0$ was an Antinode?

To truly master this concept, let's explore a variation. What if the problem had specified that $x=0$ must be an antinode?

An antinode is a point of maximum constructive interference, where the displacement reaches its maximum possible value of $\pm 2a$. For this to happen, the two waves must interfere constructively at the origin:

Substituting our expressions at $x=0$:

This condition is satisfied when the waves are perfectly in phase, which means: 1. Amplitude: $A=a$ 2. Phase Constant: $\varphi =0$

Substituting these into our general equation for $y_2$ gives:

This simple exercise shows how a small change in the boundary condition (node vs. antinode) completely changes the phase of the reflecting wave. Keep this in mind, as such variations are highly common in competitive exams like JEE Advanced!