Which of the following wave functions satisfies the wave equation, Eq. (15.12)? (a) y(x, t)=A co

Which of the following wave functions satisfies the wave...

Key Concepts

Wave Equation

The wave equation is a second-order partial differential equation, generally written as ?²y/?t² = v² ?²y/?x², that describes how waves propagate through a medium. It requires that any valid wave function exhibits specific relationships between its spatial and temporal second derivatives, ensuring that disturbances travel at a constant speed v.

Sinusoidal Wave Solutions

Sinusoidal functions, such as sine and cosine with combined phase arguments (e.g., kx ± ?t), are typical solutions to the wave equation. These functions inherently satisfy the necessary differential properties, provided that the angular frequency ? and the wave number k obey a particular relationship, making them ideal models for traveling waves.

Dispersion Relation

In the context of sinusoidal wave solutions to the wave equation, the dispersion relation (?² = v²k²) connects the wave’s spatial and temporal characteristics. This relation is essential to ensure that the chosen wave function correctly represents a propagating wave with speed v, allowing one to verify whether a function is a valid solution of the wave equation.

Transverse Motion Derivatives

For waves where the medium oscillates perpendicularly to the direction of propagation (transverse waves), the transverse velocity and acceleration of a particle at a fixed position are found by taking the first and second partial derivatives of the displacement function with respect to time. These derivatives provide insight into the dynamic response of individual particles in the medium as the wave passes through.

Transcript

00:01 In this question, we have three wave functions that we need to check if they satisfy the wave equation here.

00:08 Now, the velocity of the wave, as we can see, most of the wave functions are giving in terms of the wave number and the angular frequency.

00:17 Therefore, we can write the speed of the wave as omega divided by k.

00:24 So this is the speed of the wave.

00:29 So once we find the second spatial derivative, and the second time derivative for the wave function, we just need to plug in the velocity here or the speed of the wave here, and then we can check if this wave function is satisfying the wave equation or not.

00:49 Now, first of all, let's start by finding the spatial derivative for this wave function.

00:55 So the special derivative is equal to minus a, k, sine, k, x plus omega t here we use the chain rule and from this if you derive this again we'll get the second special derivative and it's equal to minus a kx square multiplied by cosine kx plus omega t which is equal to minus k square y of x and which is the original wave function.

01:43 Next let's find the time derivative.

01:46 So partial y, partial x, partial t is equal to a minus a omega sine kx plus omega t and if we derive this again, we'll get the second time derivative which is equal to minus a omega square cosine sign kx plus omega t which is again the original wave function multiplied by negative omega square now if we take this quantity here and multiply it with 1 over v square what we will get is minus omega square y x and t multiplied by k square divided by omega square these two terms will cancel and we will get minus k square y x and t which is equal to the second spatial derivative so we can say this wave function satisfy the wave equation the second wave function we have here is y of x and t equal to a sign k x plus omega t and we'll start again by finding the spatial so partial y, partial x is equal to k a cosine kx plus omega t.

03:38 And from this, the second derivative, the second derivative is minus k square a sine kx plus omega t, which is equal to minus k square y x and x.

03:58 And t.

03:59 So this is again the original wave function...

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