Derivation of the One-Dimensional Wave Equation - 2.4 | Theme C: Wave Behaviour | IB MYP Grade 11 Physics
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Interactive Audio Lesson

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Introduction to Wave Equation Derivation

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0:00
Teacher
Teacher

Today, we'll derive the one-dimensional wave equation for a uniform string under tension. Let’s start by understanding what we mean by a wave equation. Can anyone tell me what it signifies in terms of wave behavior?

Student 1
Student 1

I think it relates to how waves travel through a medium.

Teacher
Teacher

Exactly! A wave equation helps us describe the displacement of a wave in both space and time. Now, let's consider a string under tension. Can anyone tell me what essential factors might affect the behavior of this string?

Student 2
Student 2

The tension in the string and its mass density!

Teacher
Teacher

Correct! We denote the tension as F_T and the linear mass density as ΞΌ. Now, imagine displacing a small segment of this string. How do you think this displacement will impact the forces acting on the segment?

Student 3
Student 3

The tension will create forces on both ends, and if it's displaced, those forces might not be equal anymore.

Teacher
Teacher

Yes! That difference in tension creates the net vertical force that we need to analyze. Let's dive deeper into this setup next.

Analyzing Forces on the String Segment

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Teacher
Teacher

Let’s express the vertical forces acting on a small segment of the string. Can someone explain how the vertical components of tension would look at a point x and at x + dx?

Student 4
Student 4

The tension at point x would have a vertical component based on the slope of the string at that point. So, we would have F_T * sin(ΞΈ) where ΞΈ represents the angle.

Teacher
Teacher

Exactly! Using the small-angle approximation, sin ΞΈ is approximately equal to βˆ‚y/βˆ‚x. Therefore, the force can be approximated as F_T * (βˆ‚y/βˆ‚x). What can we say about the vertical component of tension at the other end, x + dx?

Student 1
Student 1

It would be F_T * (βˆ‚y/βˆ‚x at x + dx).

Teacher
Teacher

Right! The net vertical force acting on the segment can then be expressed as the difference between these two components. By setting this equal to the mass times acceleration, what do we form?

Student 2
Student 2

We form a differential equation that relates the acceleration to the displacement of the string!

Teacher
Teacher

Precisely! Let's write that out mathematically as we approach the wave equation.

Deriving the Wave Equation

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Teacher
Teacher

Now, let's apply Newton’s second law. The mass of the small segment is ΞΌ * dx and the vertical acceleration can be expressed as βˆ‚Β²y/βˆ‚tΒ². Using these inputs, our equation becomes ΞΌ dx * βˆ‚Β²y/βˆ‚tΒ² = F_T * βˆ‚Β²y/βˆ‚xΒ² dx. What can we simplify next?

Student 3
Student 3

We can cancel dx from both sides, leading to ΞΌ βˆ‚Β²y/βˆ‚tΒ² = F_T * βˆ‚Β²y/βˆ‚xΒ².

Teacher
Teacher

Correct! Now, if we rearrange this, what form do we get?

Student 4
Student 4

We get βˆ‚Β²y/βˆ‚xΒ² = (ΞΌ/F_T) * βˆ‚Β²y/βˆ‚tΒ².

Teacher
Teacher

Well done! This is our one-dimensional wave equation. It shows us how the spatial derivative of displacement relates to the temporal derivative. Why is this significant?

Student 1
Student 1

It helps describe how waves propagate through the string and is fundamental in wave mechanics!

Teacher
Teacher

Exactly! This equation is key in understanding wave phenomena. Let’s conclude our session with a look at what general solutions look like.

General Solution of the Wave Equation

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Teacher
Teacher

The general solution to our one-dimensional wave equation can be expressed as y(x, t) = f(x - vt) + g(x + vt). Can anyone explain what this means?

Student 2
Student 2

It means we can have two waves traveling in opposite directions along the string!

Teacher
Teacher

Yes! This indicates that the total displacement is the superposition of both waves. What does this imply for the various types of waves we can represent?

Student 3
Student 3

It allows us to represent a wide variety of waveforms using basic sinusoidal functions.

Teacher
Teacher

That's right! For example, sinusoidal waves can be represented as y(x, t) = A * sin(kx - Ο‰t + Ο†), where we incorporate amplitude, wave number, and angular frequency into the solution. Remember, both f and g can be any twice-differentiable function. This wraps up our discussion on the derivation of the wave equation and its implications!

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section describes the derivation of the one-dimensional wave equation for a uniform string under tension.

Standard

In this section, we explore how to derive the one-dimensional wave equation from basic principles involving tension and mass density of a string. We analyze the forces acting on a small element of the string as it undergoes transverse oscillations and derive the mathematical relationship governing wave propagation.

Detailed

In the derivation of the one-dimensional wave equation, we begin with a uniform string subjected to a tension force. The string's linear mass density is noted, and we consider a small segment of the string displaced vertically by a small amount. By examining the vertical forces acting on this segment, we account for the tension acting tangentially at both ends. The curvature of the string introduces a vertical component of the tension, leading to a net vertical force that results from the difference in slopes at the ends of the segment.

Next, we apply Newton's second law to relate the mass of the segment to its acceleration based on this net force. This leads to a partial differential equation that describes how any small transverse perturbation must behave. By rearranging and adjusting the equation, we arrive at the one-dimensional wave equation, which connects spatial and temporal changes in displacement of the wave on the string. This equation is essential in wave mechanics, showing how waves travel through a medium and leads us to the general solution of the wave equation, illustrating both forward- and backward-traveling waves.

Audio Book

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Setup for the Derivation

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Consider a uniform string under tension F_T with linear mass density ΞΌ. Let us derive the partial differential equation that any small transverse perturbation must satisfy.

  1. Setup.
  2. Let the string lie along the x-axis in equilibrium. Displace a small segment centered at x by an amount y(x,t).
  3. Consider an infinitesimal element of string between x and x+dx.

Detailed Explanation

In this chunk, we introduce the setup for deriving the one-dimensional wave equation. We start by considering a uniform string under tension, which means the string is taut and can transmit waves. We denote the linear mass density of the string as ΞΌ, which describes how much mass there is along a unit length of the string. The string lies along the x-axis and has a small segment that we displace vertically by an amount y(x,t). This displacement represents a small perturbation in the string's shape, which can travel as a wave.

We also highlight an infinitesimal section of the string between the points x and x + dx. This small segment will allow us to apply Newton's second law and derive the equations governing wave motion.

Examples & Analogies

Imagine plucking a guitar string. When you pull it down and let it go, you are creating a small perturbation in the string's equilibrium position. This example can help relate to how any small displacement on a string forms the foundation for wave transmission.

Vertical Forces on the Element

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  1. Vertical Forces on the Element.
  2. The vertical component of tension at x is F_T sin ΞΈ(x) β‰ˆ F_T βˆ‚y/βˆ‚x (small-angle approximation).
  3. At x + dx, the vertical component is F_T sin ΞΈ(x + dx) β‰ˆ F_T βˆ‚y/βˆ‚x(x + dx).
  4. The net vertical force Ξ”F_y acting on the element is: Ξ”F_y = F_T[βˆ‚y/βˆ‚x(x + dx) βˆ’ βˆ‚y/βˆ‚x(x)] β‰ˆ F_T βˆ‚Β²y/βˆ‚xΒ² dx.

Detailed Explanation

In this chunk, we analyze the forces acting on the small segment of the string that has been displaced. The tension in the string exerts a force that has both vertical and horizontal components. We only consider the vertical component because it causes the displacement of the string. Using the small-angle approximation (the angles are so small that sine can be approximated by the tangent), we find expressions for the vertical components of tension at positions x and x + dx.

The difference between these two forces gives us the net vertical force acting on the segment. This horizontal balance leads us to derive the second derivative of y with respect to x, which is crucial for forming the wave equation.

Examples & Analogies

Think of the string segment as a mini trampoline section. When you step on one side, it dips down, and the tension in the trampoline fabric pulls upwards. The force you feel on the trampoline is caused by the tension trying to return to its equilibrium position, just like the string tries to return when displaced.

Newton’s Second Law for the Element

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  1. Newton’s Second Law for the Element.
  2. The mass of the element is dm = ΞΌ dx. Its vertical acceleration is βˆ‚Β²y/βˆ‚tΒ².
  3. Therefore, ΞΌ dx βˆ‚Β²y/βˆ‚tΒ² = F_T βˆ‚Β²y/βˆ‚xΒ² dx.
  4. Canceling dx gives the one-dimensional wave equation: βˆ‚Β²y/βˆ‚xΒ² = (ΞΌ/F_T) βˆ‚Β²y/βˆ‚tΒ².

Detailed Explanation

Here, we apply Newton's Second Law, which states that the force acting on an object is equal to the mass times its acceleration (F = ma). We first determine the mass of our small piece of the string using the linear mass density ΞΌ times the length dx. The vertical acceleration of this element of string corresponds to the change in displacement over time, given as the second derivative of y with respect to time.

We set up our equation using the net force from the vertical components of tension and relate it to the acceleration by Newton's law. Dividing through by the section length dx yields our primary result: the one-dimensional wave equation. This equation relates how the shape of the wave changes in space to its changes over time, which is the foundational equation for describing wave behavior.

Examples & Analogies

Visualize riding on a roller coaster: as you climb higher, you're pulled down by gravity (the force). Your mass times the acceleration from the drop will determine how fast you go down. This is analogous to how forces and mass work together to create waves on the string.

Simplifying to Wave Equation

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  1. General Solution.
  2. The general solution in one dimension is any function of the form y(x, t) = f(x βˆ’ vt) + g(x + vt), where f and g are arbitrary twice-differentiable functions.
  3. For a sinusoidal (monochromatic) wave traveling in the +x direction, y(x, t) = A sin(k x βˆ’ Ο‰ t + Ο†), with Ο‰/k = v.

Detailed Explanation

The final chunk addresses the general solutions of the derived wave equation. We establish that the solution to our one-dimensional wave equation can be expressed as the sum of two waveforms moving in opposite directions. This concept is crucial because it shows that waves can travel both ways along the same medium.

For specific cases, such as monochromatic waves (waves of a single frequency), we further refine our equation to represent a sine wave traveling in one direction. This notation includes terms for amplitude, wave number, angular frequency, and phase, all of which describe the wave's characteristics.

Examples & Analogies

Imagine throwing two stones into a still pond, one from each side. Each stone creates its own waves, which travel outward. The combination of these waves creates a pattern on the surface of the water, much like how our derived wave equation shows two waves interacting along a string.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Tension (F_T): The force in the string that provides the restoring action for oscillations.

  • Linear Mass Density (ΞΌ): The mass per unit length of the string which affects wave speed.

  • Wave Equation: The equation βˆ‚Β²y/βˆ‚xΒ² = (ΞΌ/F_T) βˆ‚Β²y/βˆ‚tΒ² describes wave propagation.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • The tension in a guitar string, when plucked, creates waves that travel down the string, allowing sound to be produced.

  • In a stretched rubber band, if you were to pluck it, the disturbance would result in waves traveling along the length of the band.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎡 Rhymes Time

  • In a string that’s tight and thin, tension’s pull makes waves begin.

πŸ“– Fascinating Stories

  • Once upon a time, in a classroom, a string was placed under tension. As students plucked it, they learned how tension affected the sound waves in ways they could measure and observe.

🧠 Other Memory Gems

  • Remember 'WAVE' for Wave equation: W for Waves travel, A for Amplitude, V for Velocity, E for Energy transfer.

🎯 Super Acronyms

'WSTR' can help

  • W: for Wave
  • S: for String
  • T: for Tension
  • R: for Resonance.

Flash Cards

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Glossary of Terms

Review the Definitions for terms.

  • Term: Wave Equation

    Definition:

    A mathematical equation that describes the propagation of waves through a medium.

  • Term: Tension (F_T)

    Definition:

    The force exerted along the length of the string that affects its oscillation.

  • Term: Linear Mass Density (ΞΌ)

    Definition:

    The mass per unit length of the string, affecting the wave speed.

  • Term: Partial Derivative

    Definition:

    A derivative taken with respect to one variable while keeping others constant.

  • Term: Displacement (y)

    Definition:

    The vertical distance moved from the equilibrium position.