Wave Equation with a Free End
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Introduction to the Wave Equation
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Today, we focus on the wave equation, \( \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} \). Can anyone tell me what this equation represents?
It describes wave motion, right?
Exactly! It's fundamental in understanding how waves propagate. Now, what happens when we have boundary conditions, especially at one end?
Is it like having one end fixed and the other free?
Yes! When one end is free, we impose conditions like \( u(0,t) = 0 \). Let's move into how we solve this using Fourier Sine Transform.
Applying Boundary Conditions
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The boundary conditions we need are \( u(0,t) = 0 \), \lim_{x \to \infty} u(x,t) = 0 \), and \frac{\partial u}{\partial t}(x,0) = 0. Why are these conditions important?
They specify how the wave behaves at certain points and times.
Exactly! Understanding these conditions helps us predict wave behavior. Let's consider the Fourier Sine Transform now. How does it help us with that?
It converts our problem into a different domain, making it easier to solve differential equations?
Absolutely right! We'll see how it simplifies the process of finding solutions.
Solving the Wave Equation
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When we apply the Fourier Sine Transform to our wave equation, we get \( \frac{\partial^2 U}{\partial t^2} = -c^2 s^2 U \). What kind of equation is this?
It looks like a second-order ordinary differential equation.
Correct! The solution to this ODE is of the form \( U(s,t) = A(s) \cos(cst) + B(s) \sin(cst) \). What can you tell me about coefficients \( A(s) \) and \( B(s) \)?
A(s) is from the initial displacement and B(s) relates to initial velocity, right?
Exactly! Since \( \frac{\partial U}{\partial t}(t,0) = 0 \), it tells us that \( B(s) = 0 \). So our solution simplifies. What are we left with?
It's just \( U(s,t) = F \{f(x)\} \cos(cst) \).
Correct! Our final solution represents how the displacement evolves over time.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
The section presents the wave equation applied to a one-dimensional medium, highlighting the boundary conditions relevant when one end is free. It explains how to solve this equation using the Fourier Sine Transform, resulting in an integral solution representing the displacement over time.
Detailed
Wave Equation with a Free End
The one-dimensional wave equation, given by \( \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} \), describes wave propagation in various physical systems. In this section, we analyze the specific case where boundary conditions are set such that the displacement at one end is free. The conditions include \( u(0,t) = 0 \) (indicating the displacement is zero at the fixed end), \( \lim_{x \to \infty} u(x,t) = 0 \) (indicating that the wave tails off), and an initial displacement condition \( u(x,0) = f(x) \) alongside an initial velocity condition \( \frac{\partial u}{\partial t}(x,0) = 0 \). To solve this problem, we apply the Fourier Sine Transform, leading to a second-order ordinary differential equation, which results in a solution expressed in terms of the initial conditions. The final solution results in a representation of how the wave propagates over time, showcasing the integral of the Fourier transform of the initial displacement multiplied by oscillating terms. This approach is essential in engineering fields for understanding behaviors of structures and materials under dynamic loading.
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Introduction to the Wave Equation
Chapter 1 of 5
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Chapter Content
Let us consider the one-dimensional wave equation:
∂²u / ∂t² = c² * ∂²u / ∂x²
With boundary conditions:
- u(0,t) = 0,
- lim (x→∞) u(x,t) = 0,
- u(x,0) = f(x),
- ∂u/∂t (x,0) = 0.
Detailed Explanation
In this section, we introduce the fundamental wave equation, which describes how waves propagate through a medium. The notation used indicates how the displacement (u) changes with respect to time (t) and space (x). Here, c represents the speed of the wave. The boundary conditions specify that at the starting point (x=0), there is no displacement. Additionally, we require that as we move infinitely far from the origin, the displacement approaches zero. The initial state of the system, given by f(x), and the rate of change of displacement at time t=0, are also defined, indicating the wave's initial conditions.
Examples & Analogies
Imagine holding one end of a rope tightly while the other end is free. If you create a wave by flicking the rope, the point where you flick it corresponds to u(0,t)=0—there's no movement at your hand. As you observe, the wave will travel along the rope but diminish as you move further away from your hand, similar to the boundary condition lim (x→∞) u(x,t)=0.
Applying the Fourier Sine Transform
Chapter 2 of 5
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Chapter Content
Apply the Fourier Sine Transform:
∂²U / ∂t² = −c²s²U
This is a second-order ODE in t.
Detailed Explanation
In this step, we apply the Fourier Sine Transform to the displacement function to convert our spatial problem into a frequency domain problem. This transforms the wave equation into a second-order ordinary differential equation (ODE) with respect to time. The term s relates to the frequency components of the original function, which are now being analyzed over time. The Fourier transform helps us manage the wave problem by simplifying it into a form that can be solved for U(s,t), which represents the transformed function in the frequency domain.
Examples & Analogies
Think of tuning a violin. The sound produced corresponds to the frequency of the strings vibrating. By applying the Fourier Sine Transform, we are effectively 'tuning' our equation, breaking down the wave into different frequency components, similar to finding the unique notes a violin string can produce.
General Solution of the ODE
Chapter 3 of 5
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Chapter Content
U(s,t) = A(s)cos(cst) + B(s)sin(cst).
Detailed Explanation
Once we have a second-order ordinary differential equation, we can express its solution in terms of two arbitrary functions, A(s) and B(s). These functions correspond to the two types of oscillatory motion: one that describes how the wave behaves in terms of cosine (A(s)) and the other in terms of sine (B(s)). This general form solves the wave equation and indicates that the wave can oscillate between different states based on these functions.
Examples & Analogies
Just like how musical notes can mix to create melodies, the wave equation's solution can produce various wave patterns through the combination of the cosine and sine functions. The sounds from a musical instrument depend on the fundamental frequency (A(s)) and its harmonics (B(s))—together they make a harmonious tune, just as the wave equation forms a complex wave profile.
Determining Initial Conditions
Chapter 4 of 5
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Chapter Content
Initial conditions give:
- U(s,0) = F{s}{f(x)} = A(s),
- ∂U/∂t (t=0) = 0 ⇒ B(s) = 0.
Detailed Explanation
To find the specific solutions A(s) and B(s), we incorporate the initial conditions into the general solution. The first condition allows us to express A(s) in terms of the Fourier transform of the original function f(x), while the second condition, stating that the initial rate of displacement is zero, leads to B(s) being determined as zero. As a result, the oscillatory behavior of the wave can be completely described using only A(s).
Examples & Analogies
Imagine starting your day lying still in bed (where your initial rate of movement is zero). When you finally get up (which corresponds to U(s,0)), your position reflects where you began (A(s)). The absence of movement at that exact moment (B(s)=0) suggests that while you are initially stationary, your motion will change based on external influences as you move through your morning routine.
Final Integral Solution
Chapter 5 of 5
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Chapter Content
Thus,
U(s,t) = F{f(x)} · cos(cst) ⇒ u(x,t) = F{f(x)} cos(cst) sin(sx) ds.
Detailed Explanation
The final result allows us to express the displacement u at position x and time t in terms of the integral of the transformed initial function F{f(x)}. Here, we see how the original function influences the solution at future times via the cosine term oscillating with time and the sine term filtering through the effects of position. This integral solution gives a comprehensive view of how the initial shape of the wave evolves over time.
Examples & Analogies
Consider throwing a pebble into a still pond. The initial splash represents your initial condition, F{f(x)}, and as the ripples spread out, the combination of traveling waves (through the cosine and sine interaction) describes how the disturbance propagates through the water, just like our calculated u(x,t) depicts how the wave moves across a medium over time.
Key Concepts
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Wave Equation: Describes how wave displacement propagates over time and space.
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Boundary Conditions: Specifications that dictate how waves behave at the ends of a medium.
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Fourier Sine Transform: A mathematical tool that assists in converting spatial variables into frequency domain equivalents.
Examples & Applications
Consider a string fixed at one end and vibrating at the other. The wave equation describes how the amplitude of vibrations changes over time.
In music, the vibration of strings can be modeled using the wave equation, showcasing how sound waves propagate from one end to the other.
Memory Aids
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Rhymes
Waves travel far, in air or in string, boundaries set limits on the motion they bring.
Stories
Imagine a tightrope walker on a stretched line, representing the displacement of waves; each step corresponds to the vibration traveling in wave motion.
Memory Tools
BFO: Boundary conditions help Frame the Origin of the wave solutions.
Acronyms
WAVE
Wave Equation is Vital for Analyzing motion in Elastic mediums.
Flash Cards
Glossary
- Wave Equation
A second-order partial differential equation that describes the propagation of waves.
- Fourier Sine Transform
A transformation that converts a function in the spatial domain into a function in the frequency domain using sine functions.
- Boundary Condition
A constraint that specifies the behavior of a physical system at the boundaries.
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