**Appendix E: Theory of Wave Motion **

### Wave motion is observed in a wide variety of systems, including water waves, vibrational waves in solids, sound waves in air, and radio and light waves. The phenomenon is closely related to simple harmonic motion, but it is a complicated subject and a full treatment is well beyond the scope of Physics 275. Most introductory physics textbooks contain a few chapters on the basic physics of waves and related phenomena. For an extended introduction to wave motion we *recommend Waves, Volume 3 of the Berkeley Physics Course, by Frank Crawford. In addition to * an excellent theoretical discussion, this book includes numerous experiments on waves that can be done at home with minimal equipment.

*To understand waves theoretically, start by considering a string with a tension T that has a * *length L and is fixed at both ends as shown in Fig. E.1. The string has a mass per unit length µ and * *therefore a total mass m=µ* *L. The equilibrium configuration is a straight line, represented by a * *dashed horizontal line in the x-direction in Figure E.1. *

*We will suppose that there is no external force on the string but it is moving. Let y(x,t) * *represent the vertical displacement of the string at point x and time t. How do we determine what * *functional form is allowed for y(x,t)? Consider a small section of displaced string, as shown in Fig * *E.2. Point P (at position x) and nearby point Q (at position x+∆x) on the string are connected by * a short section of string. Since string is flexible, the two points are not rigidly connected and the section can bend. The acceleration of the section of string is controlled by the total force on it and these forces are due to the tension in the string. If we look at the small section of string between *P and Q, the tension T produces forces on the ends of the string that have equal magnitude T and * are in almost opposite directions (see Fig. E.2). The key thing to realize is that if there is some curvature in the string, the force at one end of the short section of string will not point exactly opposite to the force on the opposite end of the short section. Thus the forces acting on the ends *of the section will not exactly cancel and there will be a non-zero net force in the y-direction acting * on the small section of string.

**Figure E.1. A string with two fixed ends is oscillating in its fundamental mode. At point P and ** *time t, the string is displaced a distance y(x,t) from equilibrium (dashed line). *

n=1

### x=L/2 y(x,t)

### x=L x

### x=0

### y

### T

### T P

### Q

### x x+ δ x

### θ (x)

### θ (x + δ x)

**Figure E.2. Detailed view of a small section of the string (red curve) between points P and Q. **

**Derivation of the Wave Equation **

### Examining Figure E.2, one can show that the y-component of the total force acting on the small section of string in the y-direction at time t is given by:

*tot*

### =

*F*

*y*

* F*

_{y}*(x+∆x) − F*

_{y}* (x) = T sin(θ(x+∆x)) − Tsin(θ(x)) * *[E.1] *

### Again, here we ignore the possibility of an external driving force acting on the wire. In Equation *E.1, the angle θ(x) is the angle that the string makes with the x-axis at point x. For small angles: *

* sin(θ) ≈ tan(θ) = "slope of string at x" = * *x* *y*

### ∆

### ∆ *. * [E.2]

### Therefore:

*x*
*x*

*x*

*y**tot*

*dx*

*T* *dy* *dx*

*T* *dy*

*F*

###

###

###

### −

###

###

###

### =

∆ +

* * [E.3]

### From Newton's second law we know that the total force acing on the small section of string in the y-direction is equal to the mass of the small section of string times the accleratin in the y- direction. Thus we can rewrite Equation [E.3] as:

###

###

###

###

###

###

###

### −

###

###

###

### =

### ∆

### =

∆

+ *x* *x*

*x*

*dx*

*dy* *dx*

*T* *dy* *dt*

*y* *x* *d* *dt*

*y*

*m* *d*

^{2}

_{2}

### µ

^{2}

_{2}

### [E.4]

*where the mass of the small section of string is m=µ∆x. Dividing both sides of Equation [E.4] by * *T∆x and taking the limit ∆x → 0, one finds: *

2 2 2 2

2

### 1

*t* *y* *v* *x*

*y*

*s*

### ∂

### = ∂

### ∂

### ∂ [E.5]

### where:

### µ

*v*

_{s}### = *T* [E.6]

*The quantity v*

*s*

### is the "wave speed" or “speed of wave propagation”. Notice also in Equation [E.5]

### that we have been careful to indicate that these are actually partial derivatives. In fact we were being a bit sloppy and Equations [E.2], [E.3] and [E.4] should also have been written as partial derivatives.

### The second order partial differential equation [E.5] is called the wave equation for a

### string. All of the functions y(x,t) that are solutions to the wave equation describe all of the possible

### ways that a string can move.

**General Solutions to the Wave Equation **

### There are some general remarks that can be made about the solutions to Equation [E.5]. In *general, ANY smooth function g whose arguments are combinations of x and t in the form x+v*

_{s}*t * *or x-v*

_{s}*t will be a solution to equation. Thus y=g(x-v*

_{s}*t) is the general solution to the equation when * *g is ANY well-behaved function of x-v*

_{s}*t. This can be shown by substituting y=g(x±v*

_{s}*t) into * *Equation [E.5] and working out the derivatives. In fact, the solution y=g(x-v*

_{s}*t) describes a wave * *that has the shape g(x) at t=0 and is moving in the positive x-direction with speed v*

s### without *changing its shape. The solution y = g(x-v*

_{s}*t) describes a wave that is moving in the negative x-* direction with speed v

s### without changing its shape.

### As we will see below, only some solutions will be consistent with the boundary conditions, *i.e. the conditions at the ends of the string where additional forces or constraints are present. *

**The Wave Speed **

### To understand why the wave speed is given by v

_{s}

### , consider the case where the *displacement of the string is of the general form g(x−v*

_{s}*t). At the point x*

_{o}* at time t*

_{o}### the *displacement is thus g(x*

_{o}*-v*

_{s}*t*

_{o}*)=g*

_{1}*. Notice that at a later time t, the point x will have exactly the * same displacement g

_{1}

### =g(x-v

_{s}

### t)=g(x

_{o}

### -v

_{s}

### t

_{o}

### ) if we choose x such that:

*x − v*

_{s}*t = x*

_{o }*− v*

_{s}*t*

*0*

### → *x = x*

_{o}*+* *v*

_{s}*(t− t*

_{o}*) * [E.7]

### Thus, the displacement that was at x

_{o}

* has moved a distance v*

_{s}*(t− t*

_{o}*) further down the string in a * *time t− t*

_{o}*. Since there is nothing special about x*

_{o}### , this is true for all points on the string. Thus the *wave is moving with a speed v*

_{s}### down the string.

**Sinusoidal Travelling Waves **

### Although in general a wave can be any well-behaved function, many properties of wave motion are most easily analyzed and understood if the wave is a sine or cosine function. For example, consider the sinusoidal travelling wave

### ) sin(

### ) ,

### ( *x* *t* *A* *kx* *t*

*y* = − ω [E.8]

### where A is the amplitude of the wave, k = 2π/λ is the wave-vector, λ is the wavelength, ω = 2πf is the angular frequency and f is the frequency. We can also write this same wave in equivalent forms as:

### [ ]

###

###

###

### −

### =

###

###

###

###

### −

### =

### −

### = *t* *A* *x* *v* *t*

*x* *k* *k* *A* *t* *kx* *A* *t* *x*

*y* λ

_{s}### π ω _{)} _{sin} ω _{sin} ^{2} sin(

### ) ,

### ( [E.9]

### The wave given by Equation [E.8] or [E.9] is a sine wave moving in the positive x direction with amplitude A and speed

### v

s### = ω/k = f λ [E.10]

### We can show that Equation [E.8] is a solution to the wave equation [E.5] by taking the appropriate derivatives. You can show that:

### ) sin(

### and )

### sin(

^{2}

_{2}

^{2}

2 2

2

*k* *A* *kx* *t*

*x* *t* *y*

*kx* *t* *A*

*y* ω ω = − − ω

### ∂

### − ∂

### −

### ∂ =

### ∂ [E.11]

### Substituting into Equation [E.5] and simplifying gives

^{2}

### µ ω

^{2}

*k =* *T* . [E.12]

### Using Equation [E.6] to eliminate T and µ, we can obtain Equation [E.10], which relates the wave

*number k and the angular frequency ω, to the wave speed v*

*s*

### .

**Standing Waves on a String with Two Fixed Ends**

### We now consider the simple situation where the string is fixed at both ends and has a total *length L. We will assume that the wave has been induced by a small oscillating force that has been * applied to the string in some way and not worry about this detail further. The travelling wave described by Equation [E.8] is a solution to the wave equation, but it cannot be an allowed solution when the wave is fixed at x =0 since being fixed means we require y = 0 at x =0 for all times t.

### Instead, the solution with two fixed ends corresponds to the superposition of two waves, one going in the positive x-direction and one going in the negative x-direction. You can think of these waves as arising from reflections of a travelling wave off of the fixed end. Consider the following addition of two such travelling waves with amplitude A/2:

### ) 2 sin(

### ) 2 sin(

### ) , ( ) , ( ) ,

### ( *x* *t* *y*

_{1}

*x* *t* *y*

_{2}

*x* *t* *A* *kx* *t* *A* *kx* *t*

*y* = + = − ω + + ω [E.13]

### Using a simple trigonometric relation, one this can be rewritten as )

### cos(

### ) sin(

### ) ,

### ( *x* *t* *A* *kx* *t*

*y* == ⋅ ω [E.14]

### This is the equation for a standing wave.

### The words “standing wave” means that the wave does not appear to be travelling along the string but staying in the same place. To see why this is what Equation [E.14] produces, note that *the x-dependence of the wave (the sin(kx) factor) is simply multiplied by a time-dependent factor * (the cos(ωt) factor). This implies that every portion of the string will appear to oscillate up and down to an extent that varies along the length of the string. The maximum amplitude of the *resulting standing wave is A at the anti-nodes (where sin(kx) = 1) . The “nodes” of the standing * wave are the locations where sin(kx) = 0, and these occur where:

* kx = 0, π, 2π, 3π, …. * [E.15]

*Since k = 2π/λ, this means the nodes are located at * 2 λ ,

*x =* *n* where n = 0, 1, 2, 3, … is a positive integer [E.16]

*Since the ends of the string are fixed, we now require the wave to have a node at x=0 and * *x=L. One then finds from Equation [E.16] that the only wavelengths that will satisfy these * **boundary conditions are **

### , 3 , 2 ,1

### 2 , =

### = *n*

*n*

*n*

*L*

### λ [E.17]

### We need to exclude the solution n = 0 since if this would give k =0 and y(x,t) = 0 for all x and t, which is the trivial solution and not a very interesting wave.

### Figure E.3 shows the first few allowed standing waves for a string of length L. From [E.14], and from the equation for the wave speed [E.6], we can determine the frequencies that will correspond to standing wave resonances on the string:

*T* *n* *v* *L*

*L* *n*

*f* *v*

_{s}*n*

*s*

###

###

###

###

### =

### =

### =

### = π λ µ

### ω

### 2 1 2

### 2 for *n* = ,1 2 , 3 , [E.18]

*The allowed frequencies are just integer multiples of the fundamental frequency f*

_{1}*=v*

_{s}*/2L. So if *

### you know the tension in the string and its’ mass per unit length, you can predict the resonant

**Figure E.3. Snapshots at fixed time showing three lowest standing waves for a string fixed at both ** ends. The fundamental mode n=1 (blue) has one anti-node (location of maximum displacement).

### The next higher mode n=2 (red) has two anti-nodes and the next higher mode n=3 (green) has three anti-nodes.

### frequencies. Of course stringed-instrument musicians to tune their instruments by adjusting the tension in the strings to give a desired pitch (where their well-tuned ears measure the frequency.) Conversely, by measuring the resonance frequencies, the length of the string, and the mass per unit length of the string, you can predict the wave speed of the string and determine its tension. This technique of measuring the frequency to find the tension is sometimes used to determine the string tension in particle and nuclear physics detectors called wire chambers.

**Standing Waves on a String with One Fixed End and one Driven End**

### The above discussion of a string with two fixed ends obviously does not apply when just one end of the string is fixed and the other end is attached to a moving speaker. This is the situation in Experiment 8 when one of the string is attached to an oscillating speaker that drives the wave motion. Equations [14] to [18] only apply to the situation where both ends are fixed, and that is not the case when a speaker is driving one end of the wire back and forth.

### To understand what happens in this case, suppose that the end of the string that is attached to the speaker is at position x = 0 along the string and that the other end of the string is held fixed at x = L. Suppose that at x=0 the speaker moves the string up and down with an amplitude of y

o### or a peak-to-peak distance 2y

o### . In the apparatus, this will be a relatively small motion (a few mm).

### On the other hand, on resonance the string will exhibit a much larger motion at the anti-nodes (typically a few cm). Suppose that on resonance, at an anti-node the string moves back and forth a *distance that is Q times larger than the drive motion, i.e. by a peak-to-peak distance 2Qy*

o### . The factor Q is called the quality factor of the resonance and if there is not much dissipation, then Q will be much larger than 1.

### For this situation, at resonance there will be a standing wave on the string of the form:

### (

_{o}

_{n}### )

*n*

*o*

*f* *t* *x* *x*

*Qy* *t* *x*

*y* ( , ) = cos( 2 π ) sin 2 π ( + ) / λ [E.19]

### where y(x,t) is the lateral displacement of the string at time t and position x along the string. At X=L, the string is fixed, so we must have at all times t:

### 0 ) 2 (

### sin ) 2 cos(

### ) ,

### ( =

###

###

### +

### =

_{o}*n*
*n*

*o*

*f* *t* *L* *x*

*Qy* *t* *L*

*y* λ

### π π [E.20]

### This can only happen at all times t if:

### 0 ) 2 (

### sin =

###

###

###

### +

_{o}*n*

*x* λ *L*

### π [E.21]

### Equation [9] is only satisfied non-trivially when:

*n* *x*

*L*

_{o}*n*

### λ π

### π _{+} _{=}

### )

### 2 ( [E.22]

### where n= 1, 2, 3, … is a positive integer. Solving for the wavelength of the n

^{th}

### mode we find:

*n* *x*

*L*

_{o}*n*

### = 2 + ( )

### λ for n = 1, 2, 3,… [E.23]

### Notice that this expression is very similar to Equation [4]. If the string was held fixed at x = 0, then x

o### would equal zero and the two expressions would be identical. You can think of x

o### as an “end correction” that arises when the string is not fixed.

### To find an expression for x

o### , recall that the end attached to the drive moves back and forth between +y

o### and -y

o### , at x = 0. Thus at x= 0 we must have a displacement of

###

###

###

### =

### =

_{o}*n*
*n*

*o*
*n*

*o*

*f* *t* *Qy* *f* *t* *x*

*y* *t*

*y* λ

### π π

### π ) cos( 2 ) sin ^{2}

### 2 cos(

### ) , 0

### ( ** ** [E.24]

### This equation can be solved for x

o### in terms of the quality factor Q:

###

###

###

### =

*x*

_{o}

^{n}### arcsin *Q* 1 2π

### λ ** ** [E.25]

### Substituting back into Equation [23] and working through some algebra, one finds:

###

###

###

###

###

###

###

### −

### = *n* *Q*

*n*

*L*

### arcsin 1 1

### 2 π

### λ for n = 1, 2, 3,… [E.26]

*Notice that if there is very little dissipation, then Q>>1, arcsin(1/Q) ~ 1/Q << 1 and Equation [26] *

### will reduce to Equation [17].

### Finally, substituting back into Equation [E.10], one can find the frequency of the nth mode is given by:

### ( )

### µ π

### µ

*T* *L*

*n* *Q* *T* *f*

_{n}*L*

### 2 / 1 arcsin 2

### 1 −

###

###

###

### = [E.27]

### This expression is very similar to Equation [E.18] for the frequency of the n

^{th}

### mode when both ends are fixed. Both expressions are linear functions of n with the same slope. The only difference is whether the intercept is zero (when the ends are fixed) or non-zero (when the end is driven). Of course this assumes that Q does not depend on n or equivalently the frequency, which may not be the case

### How big will this intercept be? Suppose the quality factor of the resonance is Q = 10, which *is not very large. Then arcsin(1/Q) ~ 0.1 and we can write: *

### ( 0 . 03 )

### 2 1 1

### 2

### 1 −

###

###

###

### ≈

###

###

###

### −

###

###

###

### ≈ *T* *n*

*L* *n* *Q*

*T*

*f*

_{n}