The transverse displacement y(x, t) of a wave on a string is given by y(x, t) = e^-(ax^2 + bt^2 + 2√

Q: Why is this considered a travelling wave and not a standing wave?

A standing wave is typically represented by a function where the spatial and temporal variables are separated, like $y(x,t) = A \sin(kx) \cos(\omega t)$. In our equation, $x$ and $t$ are bound together in the form $(kx + \omega t)$, which is the hallmark of a travelling wave.

Q: Why does a positive sign in $(kx + \omega t)$ mean the wave travels in the negative x-direction?

Think about tracking a specific point on the wave, like its peak. For the peak to maintain its state, the phase $(kx + \omega t)$ must remain constant. As time $t$ increases, the only way for the sum to stay constant is if $x$ decreases. Hence, the wave moves in the negative $x$-direction.

Q: What does the shape of this wave look like physically?

Because the function is an exponential decay of a squared term ($e^{-z^2}$), it forms a bell-shaped curve known as a Gaussian pulse. It's a single, localized bump moving along the string, rather than a continuous series of ripples.

Q: How do we know that the speed is $\omega/k$?

If you track a point of constant phase, $kx - \omega t = \text{constant}$. Differentiating this with respect to time $t$ gives $k \frac{dx}{dt} - \omega = 0$. Since $\frac{dx}{dt}$ is the velocity $v$, we get $v = \frac{\omega}{k}$.

Q: Can any function of $(kx \pm \omega t)$ be a wave?

Yes! As long as the function is twice-differentiable, any function $f(kx \pm \omega t)$ is a valid mathematical solution to the linear wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$.

Question

The transverse displacement $y(x, t)$ of a wave on a string is given by
$y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$.
This represents a

Options:

(A) wave moving in $-x$ direction with speed $\sqrt{\frac{b}{a}}$

(B) standing wave of frequency $\sqrt{b}$

(C) standing wave of frequency $1/\sqrt{b}$

(D) wave moving in $+x$ direction with speed $\sqrt{\frac{a}{b}}$

Sigma Percentile · Accepted Answer

### Unmasking the Wave Equation

When we first look at the given equation, $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$, it might seem a bit intimidating. It doesn't look like the familiar sine or cosine waves we are used to seeing in textbooks. Instead, it features an exponential function. However, in physics, a wave doesn't strictly have to be sinusoidal. Any function that maintains its shape while propagating through space can be considered a travelling wave.

To determine if this mathematical expression represents a travelling wave, we need to check if it can be molded into the standard form of a wave equation, which is $y = f(kx \pm \omega t)$. Here, $f$ can be any twice-differentiable function, $k$ is the wave number, and $\omega$ is the angular frequency.

### The Magic of the Perfect Square

The secret to unlocking this problem lies in the exponent of the natural number $e$. Let's isolate and examine the expression: $ax^2 + bt^2 + 2\sqrt{ab}xt$.

If you look closely, this is a classic algebraic identity in disguise. It perfectly matches the expansion of $(A + B)^2 = A^2 + B^2 + 2AB$. By setting $A = \sqrt{a}x$ and $B = \sqrt{b}t$, we can elegantly condense the entire exponent into a perfect square:

$$ax^2 + bt^2 + 2\sqrt{ab}xt = (\sqrt{a}x + \sqrt{b}t)^2$$

Substituting this beautifully simplified expression back into our original equation, we get:

$$y(x, t) = e^{-(\sqrt{a}x + \sqrt{b}t)^2}$$

### Decoding Direction and Speed

Now, our equation is in the perfect form of $f(kx + \omega t)$, where the function $f(z) = e^{-z^2}$. This specific shape is known as a Gaussian pulse—a single, localized bump travelling along the string.

**Direction:** The sign between the spatial term ($x$) and the temporal term ($t$) dictates the direction of propagation. Because we have a positive sign ($+$) between $\sqrt{a}x$ and $\sqrt{b}t$, the wave must be travelling in the **negative $x$-direction**. If it were a negative sign, the wave would be moving to the right.

**Speed:** The speed $v$ of any travelling wave is the ratio of the coefficient of $t$ to the coefficient of $x$. In our standard form, this is $v = \frac{\omega}{k}$.

By comparing our equation $e^{-(\sqrt{a}x + \sqrt{b}t)^2}$ with the standard form, we can easily identify:
- $k = \sqrt{a}$
- $\omega = \sqrt{b}$

Therefore, the speed of the wave is:

$$v = \frac{\sqrt{b}}{\sqrt{a}} = \sqrt{\frac{b}{a}}$$

In conclusion, the given mathematical expression represents a wave pulse moving in the $-x$ direction with a speed of $\sqrt{\frac{b}{a}}$.

Animated Solution for Physics - Waves: The transverse displacement $y (x, t)$ of a wave on a string is given by $y (x, t) = e^{- (a x^{2} + b t^{2} + 2 ab x t)}$ . This represents a

Select Answer:

Visualized Solution

Analyzing the Given Equation

Standard Form of a Travelling Wave

Completing the Perfect Square

Rewriting the Wave Equation

Direction of Propagation

Wave Speed Formula

Calculating the Final Speed

The Sigma Insight: Wave Equation and Wave Speed

Unmasking the Wave Equation

The Magic of the Perfect Square

Decoding Direction and Speed

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Animated Solution for Physics - Waves: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The transverse displacement y(x,t) of a wave on a string is given by y(x,t)=e−(ax2+bt2+2ab​xt). This represents a