The amplitude of a wave disturbance travelling in the positive x-direction is given by y = 1/(1+x^2)

Q: Why does replacing $x$ with $x - vt$ represent a wave moving in the positive $x$-direction?

Replacing $x$ with $x - vt$ shifts the entire graph of the function to the right by a distance of $vt$. As time $t$ increases, this shift increases linearly, representing a wave profile moving steadily in the positive $x$-direction.

Q: What was the typo in the original JEE 1990 paper?

The paper printed the $t=0$ profile as $y = 1/(1+x)^2$ instead of $y = 1/(1+x^2)$. This typo introduced an asymptote at $x = -1$ for $t=0$, which did not exist in the $t=2\text{ s}$ profile, implying an impossible change of shape.

Q: Can a wave change its shape as it travels in reality?

Yes, in dispersive media, different frequency components of a wave travel at different speeds, causing the wave pulse to spread out or change shape. However, in introductory physics, we assume non-dispersive media where the shape remains perfectly intact.

Q: How do we find the peak of the wave from the equation?

To maximize the fraction $y = 1/(1+X^2)$, we must minimize the denominator. The minimum value of $1+X^2$ occurs when $X = 0$. Thus, the peak always occurs where the squared term in the denominator becomes zero.

Q: Is the wave velocity always constant?

Yes, the velocity of a wave is determined solely by the properties of the medium (like tension and mass density for a string, or elasticity and density for sound) and remains constant as long as the medium is uniform.

Question

The amplitude of a wave disturbance travelling in the positive $x$-direction is given by $y = \frac{1}{1+x^2}$ at time $t = 0$ and by $y = \frac{1}{1+(x-1)^2}$ at $t = 2	ext{ s}$, where $x$ and $y$ are in metre. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is ...... m/s.

Sigma Percentile · Accepted Answer

### The Magic of Wave Translation

Imagine a wave pulse traveling along a string. If the medium is non-dispersive, the wave preserves its shape perfectly as it glides through space.

How do we represent this mathematically?

If a static shape is described by a function $y = f(x)$ at $t = 0$, then moving this shape to the right with a constant velocity $v$ simply means shifting its coordinate system.

At any later time $t$, the shape has traveled a distance $\Delta x = vt$.

Therefore, the new profile is given by replacing $x$ with $x - vt$:

$$y(x, t) = f(x - vt)$$

This simple yet profound translation principle is the key to unlocking this classic JEE problem.

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### Analyzing the Setup

We are given the wave profile at two distinct moments in time:

1. At $t = 0$:
   $$y(x, 0) = \frac{1}{1+x^2}$$
   This is a bell-shaped curve (a Lorentzian function) centered at $x = 0$.

2. At $t = 2	ext{ s}$:
   $$y(x, 2) = \frac{1}{1+(x-1)^2}$$
   This is the exact same bell-shaped curve, but now centered at $x = 1$.

Notice how the peak of the wave has shifted from $x = 0$ to $x = 1$ over a time interval of $\Delta t = 2	ext{ s}$.

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### Calculating the Velocity

Let's apply our translation formula. Since the wave travels in the positive $x$-direction with velocity $v$, the general equation of the wave is:

$$y(x, t) = \frac{1}{1+(x-vt)^2}$$

Now, let's substitute $t = 2	ext{ s}$ into our general equation:

$$y(x, 2) = \frac{1}{1+(x-2v)^2}$$

We compare this derived profile with the given profile at $t = 2	ext{ s}$:

$$\frac{1}{1+(x-2v)^2} = \frac{1}{1+(x-1)^2}$$

For these two expressions to be identical for all values of $x$, their denominators must be equal:

$$x - 2v = x - 1$$

Solving this simple linear equation:

$$2v = 1 \implies v = 0.5	ext{ m/s}$$

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### The Famous Typo of 1990

If you look at the original printed paper of JEE 1990, you will find an asterisk next to this question in many answer keys. Why?

Because of a printing error, the equation at $t = 0$ was written as:

$$y = \frac{1}{(1+x)^2}$$

while the equation at $t = 2	ext{ s}$ was written as:

$$y = \frac{1}{1+(x-1)^2}$$

Mathematically, the first function has an asymptote (it goes to infinity at $x = -1$), whereas the second function is defined and finite everywhere. This means the wave **changed its shape** during propagation, violating the core premise of the question!

With the standard correction of the first equation to $y = \frac{1}{1+x^2}$, the problem becomes beautifully consistent, yielding the elegant answer of **$0.5	ext{ m/s}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Wave Pulse at $t = 0$

The Mathematical Form of a Travelling Wave

Formulating the General Wave Equation

Substituting $t = 2 s$ into the General Equation

Comparing with the Given Wave Profile at $t = 2 s$

Solving for Wave Velocity $v$

The Story of the Disputed Typo

The Sigma Insight: Wave Equation and Wave Speed

The Magic of Wave Translation

Analyzing the Setup

Calculating the Velocity

The Famous Typo of 1990

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