A progressive wave travelling along the positive x-direction is represented by y(x,t) = A sin(kx - ω

Q: Why can't the phase constant $\phi$ just be $0$ since the wave starts at the origin?

While it's true that $\sin(0) = 0$, which matches the origin point, the phase constant also dictates the initial direction of the wave. If $\phi = 0$, the wave equation at $t=0$ becomes $y = A \sin(kx)$. For small positive $x$, this gives a positive $y$ (the wave goes up). However, the given graph shows $y$ going negative initially (the wave goes down). Therefore, $\phi$ cannot be $0$.

Q: What if the wave in the graph started by going upwards instead of downwards?

If the wave went upwards initially, then for a small positive $x$, the displacement $y$ would be positive. In that specific scenario, testing $\phi = 0$ would yield $y = A \sin(kx)$, which is positive and matches the graph. So, if the wave went up first, $\phi = 0$ would be the correct answer.

Q: Can we use the mathematical slope (derivative) to find $\phi$ instead of just looking at the graph?

Absolutely! The slope $\frac{dy}{dx}$ of the snapshot $y = A \sin(kx + \phi)$ is $Ak \cos(kx + \phi)$. At the origin ($x=0$), the slope is $Ak \cos(\phi)$. From the visual graph, the slope at the origin is clearly negative. Since $A$ and $k$ are positive, $\cos(\phi)$ must be negative. We already knew $\sin(\phi)=0$, so $\phi$ must be $\pi$ because $\cos(\pi)=-1$ (which is negative), whereas $\cos(0)=1$ (which is positive).

Q: Does the fact that the wave is travelling in the positive x-direction affect the snapshot?

The snapshot is taken at a fixed, frozen moment in time ($t=0$). The direction of travel (indicated by the negative sign in $-\omega t$) determines how the wave will move and evolve as time progresses *after* $t=0$. However, it does not change the static shape of the snapshot itself at that exact instant.

Q: What does the phase constant $\phi$ actually represent in physical terms?

The phase constant $\phi$ is like the 'starting configuration' of the wave. It determines the exact position and the initial direction of oscillation for the particle located at the origin ($x=0$) at the very beginning of our observation ($t=0$).

Question

A progressive wave travelling along the positive x-direction is represented by $y(x,t) = A \sin(kx - \omega t + \phi)$. Its snapshot at $t = 0$ is given in the figure.

Options:

(A) $-\frac{\pi}{2}$

(B) $\pi$

(C) $0$

(D) $\frac{\pi}{2}$

Sigma Percentile · Accepted Answer

### The Anatomy of a Wave

Imagine standing by a calm lake and tossing a pebble into the water. The ripples that travel outward are a perfect physical manifestation of a progressive wave. In physics, we describe this elegant motion using a mathematical equation that captures both space and time.

The general equation for a progressive wave travelling in the positive $x$-direction is given by:

$$y(x,t) = A \sin(kx - \omega t + \phi)$$

Let's break down this beautiful equation. The variable $y$ represents the displacement of a particle from its mean position. $A$ is the amplitude, the maximum height the wave reaches. The term $kx$ tells us how the wave varies in space, where $k$ is the wave number. The term $\omega t$ tells us how the wave oscillates in time, where $\omega$ is the angular frequency.

But what about $\phi$? This is the **phase constant** or initial phase. It is the crucial piece of information that tells us exactly where the wave cycle begins at $x=0$ and $t=0$. Finding this $\phi$ is the core objective of our problem.

### Freezing Time: The Snapshot

The problem provides us with a "snapshot" of the wave at $t = 0$. Think of this as taking a photograph of the lake exactly at the moment your stopwatch starts.

Mathematically, taking a snapshot at $t=0$ means we substitute $t=0$ into our wave equation.

$$y(x,0) = A \sin(kx - \omega(0) + \phi)$$

$$y(x,0) = A \sin(kx + \phi)$$

This simplified equation now describes the shape of the wave frozen in space, exactly as shown in the provided graph.

### The Origin Story: Boundary Conditions

To find the unknown phase constant $\phi$, we need to act like detectives and look for clues in the graph. The most obvious clue is right at the origin, where $x = 0$.

Looking at the graph, we can clearly see that the wave passes exactly through the origin. This means that at $x = 0$, the displacement $y$ is also $0$. Let's plug this boundary condition into our snapshot equation:

$$y(0,0) = A \sin(k(0) + \phi) = 0$$

$$A \sin(\phi) = 0$$

Since the amplitude $A$ is not zero (otherwise there would be no wave!), it must be that $\sin(\phi) = 0$.

From our knowledge of trigonometry, we know that the sine function is zero at multiple angles. The two most common principal values are:

$$\phi = 0 \quad 	ext{or} \quad \phi = \pi$$

We have narrowed down the possibilities, but we are not done yet. We need a tie-breaker.

### The Decisive Slope: Breaking the Tie

How do we choose between $\phi = 0$ and $\phi = \pi$? The secret lies in the **direction** the wave is heading right after the origin.

Look closely at the graph just to the right of the y-axis (for small positive values of $x$). The wave curve dips downwards below the x-axis. This means that for a small $x > 0$, the displacement $y$ must be **negative**.

Let's test our two candidate values for $\phi$ to see which one produces this negative displacement.

**Case 1: What if $\phi = 0$?**

If we substitute $\phi = 0$ into our snapshot equation, we get:

$$y(x,0) = A \sin(kx + 0) = A \sin(kx)$$

For a small positive $x$, the value of $kx$ is a small positive angle in the first quadrant. The sine of a first-quadrant angle is positive. Therefore, $y$ would be positive. This would mean the wave goes *upwards* from the origin. But our graph clearly shows it going downwards! So, $\phi = 0$ is incorrect.

**Case 2: What if $\phi = \pi$?**

Let's substitute $\phi = \pi$ into our snapshot equation:

$$y(x,0) = A \sin(kx + \pi)$$

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

$$y(x,0) = -A \sin(kx)$$

Now, for a small positive $x$, $\sin(kx)$ is positive, but the negative sign in front makes the entire expression for $y$ **negative**. This perfectly matches our visual observation of the wave dipping below the x-axis!

### The Final Verdict

By carefully analyzing both the position of the wave at the origin and its initial slope, we have conclusively proven that the phase constant must be $\pi$.

This elegant interplay between the visual geometry of a graph and the rigorous algebra of trigonometric functions is what makes wave mechanics so deeply satisfying to study. The correct option is indeed **(b)**.

Animated Solution for Physics - Waves: A progressive wave travelling along the positive x-direction is represented by $y (x, t) = A sin (k x - ω t + ϕ)$ . Its snapshot at $t = 0$ is given in the figure.

Select Answer:

Visualized Solution

$y (x, 0)$

$y (0, 0) = 0$

$y < 0$ for $x > 0$

$ϕ = π$

The Sigma Insight: Wave Equation and Wave Speed

The Anatomy of a Wave

Freezing Time

The Origin Story

The Decisive Slope

The Final Verdict

Similar Questions

A wave is represented by the equation; $y = A sin (10 π x + 15 π t + π /3)$ where, $x$ is in metre and $t$ is in second. The expression represents

A wave equation which gives the displacement along the $y$ -direction is given by : $y = 1 0^{- 4} sin (60 t + 2 x)$ where, $x$ and $y$ are in metre and $t$ is time in second. This represents a wave

The displacement $y$ of a particle in a medium can be expressed as $y = 1 0^{- 6} sin (100 t + 20 x + \frac{π}{4}) m$ where, $t$ is in second and $x$ in metre. The speed of the wave is

A sound wave of frequency $245 Hz$ travels with the speed of $300 ms^{- 1}$ along the positive X-axis. Each point of the wave moves to and fro through a total distance of $6 cm$ . What will be the mathematical expression of this travelling wave?

The displacement $y$ of a wave travelling in the $x$ -direction is given by $y = 1 0^{- 4} sin (600 t - 2 x + \frac{π}{3})$ metre where, $x$ is expressed in metres and $t$ in seconds. The speed of the wave-motion, in $ms^{- 1}$ is

A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of $3 m/s$ and $90 m/s^{2}$ respectively. Velocity of the wave is $20 m/s$ . Find the waveform.