Domanda del libro di testo
What is the frequency of an electromagnetic wave that has the same wavelength as a 2.5 kHz sound wave in water?
472
views
Ch 16: Traveling Waves
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
Non è quello che usi tu?Cambia libro di testoScegli un libro di testo diverso
TOC.tableOfContents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 16, Problema 15
Show that the displacement D(x,t) = cx² + dt², where c and d are constants, is a solution to the wave equation. Then find an expression in terms of c and d for the wave speed.
Guida verificata passo dopo passo1
Step 1: Recall the wave equation, which is typically written as ∂²D/∂t² = v² ∂²D/∂x², where D(x,t) is the displacement, v is the wave speed, and ∂²D/∂t² and ∂²D/∂x² are the second partial derivatives of D with respect to time and position, respectively.
Step 2: Compute the second partial derivative of D(x,t) = cx² + dt² with respect to time (t). Since the term cx² does not depend on t, its derivative with respect to t is zero. For the term dt², the second derivative with respect to t is 2d.
Step 3: Compute the second partial derivative of D(x,t) = cx² + dt² with respect to position (x). Since the term dt² does not depend on x, its derivative with respect to x is zero. For the term cx², the second derivative with respect to x is 2c.
Step 4: Substitute the computed second derivatives into the wave equation ∂²D/∂t² = v² ∂²D/∂x². This gives 2d = v² * 2c. Simplify this equation to find the relationship between v, c, and d.
Step 5: Solve for the wave speed v in terms of c and d. From the equation 2d = v² * 2c, divide both sides by 2c to isolate v², yielding v² = d/c. Take the square root of both sides to find v = √(d/c).
Risposta video verificata per un problema simile:
Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
3mConcetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Wave Equation
The wave equation is a second-order partial differential equation that describes the propagation of waves through a medium. It is typically expressed as ∂²D/∂t² = v²∂²D/∂x², where D is the displacement, t is time, x is position, and v is the wave speed. Understanding this equation is crucial for analyzing wave behavior and determining if a given function, like D(x,t), satisfies the conditions of wave propagation.
Video consigliato:
Percorso guidato
06:28
Equations for Transverse Standing Waves
Displacement Function
In the context of wave mechanics, the displacement function D(x,t) represents the position of points in a medium as a function of both space (x) and time (t). The form D(x,t) = cx² + dt² indicates that the displacement is a combination of spatial and temporal components, where c and d are constants. Analyzing this function helps in verifying if it meets the criteria set by the wave equation.
Video consigliato:
Percorso guidato
08:30Intro to Wave Functions
Wave Speed
Wave speed is the rate at which a wave propagates through a medium and is denoted by v in the wave equation. It can be derived from the relationship between the second derivatives of the displacement function with respect to time and space. In this case, finding the wave speed involves substituting the displacement function into the wave equation and solving for v in terms of the constants c and d.
Video consigliato:
Percorso guidato
07:19Intro to Waves and Wave Speed
Pratica correlata
Domanda del libro di testo
Show that the displacement D(x,t) = ln(ax + bt), where a and b are constants, is a solution to the wave equation. Then find an expression in terms of a and b for the wave speed.
1682
views
Domanda del libro di testo
A hammer taps on the end of a 4.00-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pulses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 9.00 ms. What is the speed of sound in this metal?
1796
views
Domanda del libro di testo
The wave speed on a string under tension is 200 m/s. What is the speed if the tension is halved?
2172
views
Domanda del libro di testo
FIGURE EX16.8 is a picture at t = 0 s of the particles in a medium as a longitudinal wave is passing through. The equilibrium spacing between the particles is 1.0 cm. Draw the snapshot graph D(x, t = 0 s) of this wave at t = 0 s.
2055
views
Domanda del libro di testo
Draw the history graph D(x = 4.0 m, t) at x = 4.0 m for the wave shown in FIGURE EX16.4.
2760
views