A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration o

Q: Why is the wave velocity different from the particle velocity?

Wave velocity is the speed at which the disturbance (energy) propagates horizontally through the medium. Particle velocity is the speed at which individual particles of the medium oscillate vertically about their equilibrium positions.

Q: How do we know whether to use a plus or minus sign in the wave equation?

The sign depends on the direction of wave propagation. A term of the form $(\omega t - kx)$ represents a wave traveling in the positive $x$-direction, while $(\omega t + kx)$ represents a wave traveling in the negative $x$-direction.

Q: What does the phase constant $\phi$ represent?

The phase constant $\phi$ describes the initial state of the wave at $t=0$ and $x=0$. It is determined by the initial conditions of the system.

Q: Can the amplitude of a transverse wave be larger than its wavelength?

Yes, mathematically there is no restriction. However, in physical strings, extremely high amplitudes relative to wavelength violate the small-angle approximation, leading to non-linear wave behavior.

Q: Why did we use partial derivatives to find particle velocity?

Because the wave displacement $y(x,t)$ depends on both position $x$ and time $t$. To find the velocity of a specific particle at a fixed position $x$, we must differentiate with respect to time $t$ while keeping $x$ constant.

Question

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

Sigma Percentile · Accepted Answer

### Introduction to Wave Mechanics

Imagine a beautifully stretched guitar string.

When plucked, a disturbance ripples across its length.

This is the magic of a **transverse wave**, where individual particles of the string dance up and down while the wave pattern itself travels horizontally.

To describe this motion mathematically, we need to bridge the gap between the microscopic motion of a single particle and the macroscopic propagation of the wave.

This problem challenges us to do exactly that by using the maximum particle velocity and acceleration to reconstruct the entire waveform.

---

### Analyzing the Setup

Let's start by defining the general equation of a progressive harmonic wave.

Any such wave can be represented as:

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

Here, $A$ is the amplitude of the wave, representing the maximum displacement of any particle from its equilibrium position.

The term $\omega$ is the angular frequency, which dictates how fast the particles oscillate in time.

The term $k$ is the wave number, describing how the wave spatial profile repeats itself over distance.

Finally, $\phi$ is the initial phase constant, determined by the boundary conditions at $t = 0$ and $x = 0$.

---

### The Kinematics of a String Particle

To find the velocity of a specific particle on the string, we take the partial derivative of the displacement $y(x, t)$ with respect to time $t$:

$$v_p = \frac{\partial y}{\partial t} = A\omega \cos(\omega t \pm kx + \phi)$$

Since the maximum value of the cosine function is $1$, the **maximum particle velocity** is:

$$v_{p,\max} = A\omega$$

Similarly, to find the acceleration of the particle, we differentiate the velocity with respect to time once more:

$$a_p = \frac{\partial^2 y}{\partial t^2} = -A\omega^2 \sin(\omega t \pm kx + \phi)$$

This shows that the particle undergoes simple harmonic motion.

The **maximum particle acceleration** is:

$$a_{p,\max} = A\omega^2$$

---

### Solving for the Wave Parameters

We are given two crucial pieces of information:

1. The maximum particle velocity is $3	ext{ m/s}$.
2. The maximum particle acceleration is $90	ext{ m/s}^2$.

Let's set up our system of equations:

$$A\omega = 3 \quad 	ext{--- (Equation 1)}$$

$$A\omega^2 = 90 \quad 	ext{--- (Equation 2)}$$

To solve for $\omega$, we can divide Equation 2 by Equation 1:

$$\frac{A\omega^2}{A\omega} = \frac{90}{3}$$

$$\omega = 30	ext{ rad/s}$$

Now, we substitute $\omega = 30	ext{ rad/s}$ back into Equation 1 to find the amplitude $A$:

$$A(30) = 3 \implies A = 0.1	ext{ m}$$

---

### Connecting Wave Speed and Wave Number

We are also given that the wave propagates with a velocity of $v_w = 20	ext{ m/s}$.

The relationship between wave velocity, angular frequency, and wave number is given by the fundamental formula:

$$v_w = \frac{\omega}{k}$$

Rearranging this to solve for the wave number $k$:

$$k = \frac{\omega}{v_w}$$

Substituting our known values:

$$k = \frac{30}{20} = 1.5	ext{ m}^{-1}$$

---

### Constructing the Final Waveform

Now we have all the ingredients to write down the complete wave equation!

Substituting $A = 0.1	ext{ m}$, $\omega = 30	ext{ rad/s}$, and $k = 1.5	ext{ m}^{-1}$ into our general equation:

$$y(x, t) = 0.1 \sin(30t \pm 1.5x + \phi)$$

If the wave is traveling in the positive $x$-direction, the sign between the time and space terms is negative:

$$y(x, t) = 0.1 \sin(30t - 1.5x + \phi)$$

If it is traveling in the negative $x$-direction, the sign is positive:

$$y(x, t) = 0.1 \sin(30t + 1.5x + \phi)$$

This elegant equation completely describes the physical state of the vibrating string at any point in space and time!

Animated Solution for Physics - Waves: 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.

Visualized Solution

Visualizing the Wave Motion

Maximum Particle Velocity Formula

Maximum Particle Acceleration Formula

Setting up the Given Values

Solving for Angular Frequency $ω$

Solving for Amplitude $A$

Relating Wave Velocity to Wave Number

Calculating Wave Number $k$

Constructing the Waveform Equation

Exploring Phase and Direction

The Sigma Insight: Wave Equation and Wave Speed

Introduction to Wave Mechanics

Analyzing the Setup

The Kinematics of a String Particle

Solving for the Wave Parameters

Connecting Wave Speed and Wave Number

Constructing the Final Waveform

Similar Questions

A transverse sinusoidal wave of amplitude $a$ , wavelength $λ$ and frequency $f$ is travelling on a stretched string. The maximum speed of any point on the string is $v /10$ , where $v$ is the speed of propagation of the wave. If $a = 1 0^{- 3} m$ and $v = 10 m/s$ , then $λ$ and $f$ are given by

A travelling wave in a stretched string is described by the equation ; $y = A sin (k x - ω t)$ The maximum particle velocity is

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 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 equation of a wave on a string of linear mass density $0.04 kg m^{- 1}$ is given by $y = 0.02 (m) sin [2 π (\frac{t}{0.04 (s)} - \frac{x}{0.50 (m)})]$ . The tension in the string is

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

Equation of travelling wave on a stretched string of linear density $5 g/m$ is $y = 0.03 sin (450 t - 9 x)$ , where distance and time are measured in SI units. The tension in the string 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 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