chapter 16: waves 1 1. the equation...a. la rosa lecture notes ph‐213 general physics chapter 16:...

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A. La Rosa Lecture Notes PH213 GENERAL PHYSICS ________________________________________________________________________ CHAPTER 16: WAVES1 1. The wave equation Outline of Mechanical waves Longitudinal and transverse waves Waves in a string, sound waves The wave equation Description of waves using functions of two variables Travelling waves The wave equation 0 y v 1 y 2 2 2 2 2 t x Waves in a string Reflection and transmission of waves at an interface

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Page 1: CHAPTER 16: WAVES 1 1. The equation...A. La Rosa Lecture Notes PH‐213 GENERAL PHYSICS CHAPTER 16: WAVES‐1 1. The wave equation Outline of Mechanical waves Longitudinal and transverse

A. La Rosa                                                    Lecture Notes 

PH‐213  GENERAL PHYSICS ________________________________________________________________________   

CHAPTER 16:  WAVES‐1 

1. The wave equation 

Outline of Mechanical waves           Longitudinal and transverse waves 

  Waves in a string, sound waves 

The wave equation   Description of waves using functions of two variables 

Travelling waves 

The wave equation   0y

v

1y2

2

22

2

tx

Waves in a string  Reflection and transmission of waves at an interface 

 

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A
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k x - w t + f
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l is the spatial periodicity
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T is the temporal periodicity
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Notice the whole expression inside the parenthesis is referred to as the phase
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Y=Y(x,t)
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Example A sinusoidal wave of frequency 500 Hz has a speed of 350 m/s. a) How far apart are two points x1 and x2 that, at a given time to,

differ in phase by /3 rad? b) What is the phase difference between two displacements Y1 and

Y2 at a certain point (at a certain xo point) at times 1 .0 ms apart? (Notice, you are not being asked for the values of Y1 nor Y2; just their relative phase difference.)

Solution

v = f f = 500 Hz and v = 350 m/s implies, = 350/500 = 0.7 m Y = A Cos (k x - t +

a) Things happen at t = t0 Phase1 = k x1 - t0 +

Phase2 = k x2 - t0 +

Phase1 - Phase2 = [k x1 - t0 + ] - [ k x2 - t0 +

= [k( x1 - x2 /3 rad = [k( x1 - x2 implies

x1 - x2 = (/3 )/k = (/3 ) /2 =

= 0.7/6 = 11.7 cm

b) Things happen at x = x0

Phase1 = k x0 - t1 +

Phase2 = k x0 - t2 +

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Phase1 - Phase2 = [ k x0 - t1 + ] - [k x0 - t2 +

= [ ( t2 - t1

= [ f ( t2 - t1

( t2 - t1 1 ms implies,

Phase1 - Phase2 = [ 500 Hz ( 10-3s

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Mechanical Waves
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Example
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The Wave Equation
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Waves on a Stretched String
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Cos = 1 – (1/2) 2 + … For small : Cos ~ 1

Sin = – (1/3!) 3 + … For small : Sin =

Tan = – (1/3!)2 3 + … For small : Tan =

For the case of the string:

Horizontal force Fx = T Cos - T Cos

Vertical force Fy = T Sin - T Sin

For the case of small and smallcos = 1, cos = 1

sin = , sin =

tan = , tan =

Horizontal force Fx = T - T = 0

Vertical force Fy = T Sin - T Sin But, conveniently for the steps to follow below, we

expressed the expression above in terms of the tangent function

= T tan - T tan

The tan can be interpreted as the slope of the y vs x curve. This is illustrated in the figure below.

Y

x

y(x,t)

Notice, at a given fixed time t:

y

x x

= tan

Geometrical interpretation

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Or,

Twhere

tx

v0

y

v

1y2

2

22

2

which is the wave equation

It admits solutions of the form

y(x,t) = f (x-vt) + g(x+vt)

where f and g are arbitrary functions

x)

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Energy content in a wave travelling along a string
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Case: y = A Sin(kx-wt) w/k = v (T/m)1/2 = v
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Propagation of energy along a string
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At x=0

The little guy A shakes his hands UP and DOWN with a (temporal) frequency f. So, at x=0 the motion is described by Y = Cos (0-t) = Cos t = 2f

At x=xo

The little guy A sees that a particular segment of the string located at x=xo (point “P”) goes UP and DOWN with frequency f. y = Cos (k xo -t)

P

f = v

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Reflection and Transmission of Waves at a Interface
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Incident
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