A transverse wave is described by the equation y = y0 sin 2π(ft - x/λ). The maximum particle velocit

Q: Why is the wave velocity constant while the particle velocity varies?

Wave velocity depends entirely on the properties of the medium (like tension and linear mass density in a string), which are constant. In contrast, individual particles of the medium undergo simple harmonic motion, meaning their velocity depends on their position and changes continuously from zero at the extremes to maximum at the mean position.

Q: How did we get $v_w = \lambda f$ from the equation?

The phase of the wave is given by $\theta = 2\pi(ft - x/\lambda)$. For a point of constant phase on the wave profile, $ft - x/\lambda = \text{constant}$. Differentiating this with respect to time gives $f - \frac{1}{\lambda}\frac{dx}{dt} = 0$, which simplifies to the wave propagation speed $v_w = \frac{dx}{dt} = \lambda f$.

Q: Why is the maximum particle velocity at the mean position?

In simple harmonic motion, a particle has maximum kinetic energy (and thus maximum speed) at the equilibrium or mean position ($y=0$), where all of its energy is kinetic. At the extreme positions ($y = \pm y_0$), the particle momentarily stops, making its velocity zero.

Q: What does the ratio of maximum particle velocity to wave velocity physically represent?

The ratio $\frac{v_{p,\text{max}}}{v_w}$ is mathematically equal to the maximum slope of the wave profile, $\left|\frac{\partial y}{\partial x}\right|_{\text{max}}$. It represents how steep the wave is.

Q: Would this relation change if the wave was travelling in the negative x-direction?

No. A wave travelling in the negative x-direction is represented by $y = y_0 \sin 2\pi(ft + x/\lambda)$. While the direction of wave propagation changes, the magnitudes of the wave velocity ($v_w = f\lambda$) and the maximum particle velocity ($v_{p,\text{max}} = 2\pi f y_0$) remain the same, so the relationship holds.

Question

A transverse wave is described by the equation $y = y_0 \sin 2\pi \left(ft - \frac{x}{\lambda}\right)$. The maximum particle velocity is equal to four times the wave velocity if

Options:

(A) $\lambda = \frac{\pi y_0}{4}$

(B) $\lambda = \frac{\pi y_0}{2}$

(C) $\lambda = \pi y_0$

(D) $\lambda = 2\pi y_0$

Sigma Percentile · Accepted Answer

Wave motion is one of the most fascinating phenomena in physics, describing how energy travels through space without the physical transport of matter.

To truly master waves, we must first shatter a common point of confusion: the difference between **wave velocity** and **particle velocity**.

Let's dive deep into this elegant problem from JEE 1984 and uncover the beautiful geometry hidden within the mathematics.

### The Two Velocities: Wave vs. Particle

Imagine a long rope stretched horizontally. If you wiggle one end up and down, a wave disturbance travels along the rope.

The speed at which this disturbance (or wave profile) moves forward is the **wave velocity** ($v_w$). For a given uniform medium under constant tension, this velocity is constant.

On the other hand, if you focus on a single red ribbon tied to one point on the rope, you will notice that the ribbon does not travel forward. Instead, it oscillates vertically up and down.

The speed of this ribbon at any instant is the **particle velocity** ($v_p$). Unlike the wave velocity, the particle velocity is constantly changing as the particle undergoes Simple Harmonic Motion (SHM).

### Deriving the Wave Velocity

We are given the wave equation:

$$y = y_0 \sin 2\pi \left(ft - \frac{x}{\lambda}ight)$$

To find the wave velocity, we can compare this with the standard progressive wave equation:

$$y = y_0 \sin(\omega t - kx)$$

By matching the coefficients, we find:

$$\omega = 2\pi f \quad 	ext{and} \quad k = \frac{2\pi}{\lambda}$$

The wave velocity ($v_w$) is defined as the ratio of the angular frequency to the wave number:

$$v_w = \frac{\omega}{k} = \frac{2\pi f}{\frac{2\pi}{\lambda}} = f \lambda$$

This is our first key piece of the puzzle.

### Finding the Maximum Particle Velocity

The particle velocity ($v_p$) is the rate of change of the vertical displacement ($y$) with respect to time ($t$) for a fixed position ($x$). Mathematically, this is the partial derivative:

$$v_p = \frac{\partial y}{\partial t}$$

Let's differentiate our wave equation with respect to $t$:

$$v_p = \frac{\partial}{\partial t} \left[ y_0 \sin 2\pi \left(ft - \frac{x}{\lambda}ight) ight]$$

Using the chain rule, we get:

$$v_p = 2\pi f y_0 \cos 2\pi \left(ft - \frac{x}{\lambda}ight)$$

Since the maximum value of the cosine function is $\pm 1$, the **maximum particle velocity** ($v_{p,	ext{max}}$) occurs when the particle passes through its equilibrium position ($y=0$):

$$v_{p,	ext{max}} = 2\pi f y_0$$

### Bringing It All Together

Now, we apply the condition given in the problem: the maximum particle velocity is equal to four times the wave velocity:

$$v_{p,	ext{max}} = 4 v_w$$

Substituting our derived expressions into this condition:

$$2\pi f y_0 = 4 (f \lambda)$$

Notice how the frequency ($f$) beautifully cancels out from both sides! This tells us that this relationship is independent of how fast we wiggle the rope; it depends purely on the spatial geometry of the wave.

$$2\pi y_0 = 4 \lambda$$

Solving for the wavelength ($\lambda$):

$$\lambda = \frac{2\pi y_0}{4} = \frac{\pi y_0}{2}$$

Thus, the correct option is **(b)**.

### The Geometric Insight

There is a profound physical meaning behind the ratio of maximum particle velocity to wave velocity. Let's look at the slope of the wave profile, which is given by the spatial derivative $\frac{\partial y}{\partial x}$:

$$\frac{\partial y}{\partial x} = -\frac{2\pi y_0}{\lambda} \cos 2\pi \left(ft - \frac{x}{\lambda}ight)$$

The maximum slope of the wave profile is:

$$\left| \frac{\partial y}{\partial x} ight|_{	ext{max}} = \frac{2\pi y_0}{\lambda}$$

Notice that:

$$\frac{v_{p,	ext{max}}}{v_w} = \frac{2\pi f y_0}{f \lambda} = \frac{2\pi y_0}{\lambda} = 	ext{Maximum Slope}$$

So, when the problem states that the maximum particle velocity is four times the wave velocity, it is geometrically stating that the maximum slope of the wave is exactly $4$! This connection between calculus, geometry, and physical wave dynamics is what makes physics truly beautiful.

Animated Solution for Physics - Waves: A transverse wave is described by the equation $y = y_{0} sin 2 π (f t - \frac{x}{λ})$ . The maximum particle velocity is equal to four times the wave velocity if

Select Answer:

Visualized Solution

Visualizing the Wave Parameters

Understanding Wave Velocity ( $v_{w}$ )

Extracting Wave Velocity from the Equation

Defining Particle Velocity ( $v_{p}$ )

Differentiating to Find Particle Velocity

Finding Maximum Particle Velocity ( $v_{p, max}$ )

Setting Up the Given Condition

Simplifying the Equation

Solving for Wavelength ( $λ$ )

Physical Significance of the Ratio

The Sigma Insight: Wave Equation and Wave Speed

The Two Velocities

Deriving the Wave Velocity

Finding the Maximum Particle Velocity

Bringing It All Together

The Geometric Insight

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