1. Harmonic Waves
Wave – defined as the motion ofdisturbance
Examples:
Mechanical wave Electromagneticwave:
- travels through a medium and
wave speed depends on the
medium
- generated by charged particles
and travels independently of
the medium
https://www.flickr.com/photos/physicsclassroom/galleries/72157625109648267/ http://www.pion.cz/en/article/electromagnetic-spectrum
When a source creates a wave that travels in simple harmonicmotion,it is
referred as a harmonic wave or sinusoidal wave.
Properties of Harmonic Waves
Amplitude: maximum displacement ofthe medium from equilibrium
position (±A)
The maximum positivedisplacement (+A)is the crest , and the
minimum displacement (-A)is the trough .
Wavelength: the distance between successive crests or between successive
troughs and it is symbolized bylambda,λ.

2. http://nothingnerdy.wikispaces.com/4.4+WAVE+CHARACTERISTICS
The vertical displacement ofan element at location x can be calculated using
the followinggeneral equation:
D(x) = A sin(kx)
Notice the new quantity wavenumber (k) in the given equation.This
quantityhas no correlation with the springconstant k! Rather,it represents
the how often a wave pattern repeats per unit distance in this context.
K =
2𝜋
λ
rad/m
Period (T): the time it takes complete one wave cycle
Frequency (f): the number of wave cycles passinga fixed point of the
medium in one second
Wave speed relationships:
An important relationship to consider for a harmonic wave is for one second,
there are f numbers of wave cycles passinga fixed point ofa medium, each
with a length λ. The wave speed is therefore: V= 𝛌𝐟 and this holds true for
any kind of periodicwaves.
Previously, we introduced wave number, k, and so 𝛌 can be rearranged as
2𝜋
k
.

3. The wave speed can be rewritten in terms of wave number and angular
frequency: v= (
2𝜋
k
)(
𝑤
2𝜋
) =
𝐰
𝐤
As time changes,the displacement ofany element of a medium changes
from its equilibrium position. Therefore, the function for a harmonicwave
can be expressed differentlydependingon whether it is travellingin the
direction of increasingx or decreasingx.
Increasingx: x= (x-vt)
Therefore, D(x) = A sin(kx)
 D(x,t) = A sin(k(x-vt))
 D(x,t) = A sin(kx – wt)
 D(x,t) = A sin (
𝟐𝝅
𝛌
x -
𝟐𝝅
𝐓
t)
*Note: kv = w, so kvt = wt
Decreasing x: D(x,t) = A sin(
𝟐𝝅
𝛌
x +
𝟐𝝅
𝐓
t)
However, notice how in the abovefunctions when the position is at x=0 and
t=0, the displacement is restricted to beingzero only.
Here is when the phase constant is added to the function to account for this
restriction.The functions can nowbe rewritten in terms of wave length and
period as well as the phase constant ϕ:
Increasingx: D(x,t) = A sin (
𝟐𝝅
𝛌
x -
𝟐𝝅
𝐓
t + ϕ )
Decreasing x: D(x,t) = A sin(
𝟐𝝅
𝛌
x +
𝟐𝝅
𝐓
t + ϕ)

4. Transverse Velocity and Acceleration
Each segment of a medium that has a harmonicwave travellingthrough it
experiences a different velocityas time changes
The velocity is the rate of change of the displacement ofthe segment,
represented by:
V(x,t)= -wA cos(kx – wt + ϕ)
Note: the speed of wave is constant through a medium, but each segment of
the medium is continuously oscillating thus changingits velocity.
Similarly,the rate of change of the instantaneous velocityofthe segment
gives acceleration
a(x,t) = -w2
A sin(kx – wt + ϕ)
Note: acceleration is proportion to its displacement, but opposite in sign.
Each segment of the medium has instantaneous acceleration,but the wave
itselfhas no acceleration.
Step by step solutions to end of chapter problems:
Problem 1:
The displacement functionat t=2.00 s for a travellingwave is given by the
followingequation:
D(x, 2.00) = 0.15m sin (10.0x-15.0)
where x is in metres and t is in seconds.
a) What are the wavelength,frequency, and speed of the wave?
b) Write a displacement equation,that is,an equation forD(x, t), for the
wave.

5. Solution:
a) Looking back at the previous notes,one way of writing the
displacement equation is: D(x,t) = A sin(kx – wt)
Notice in the question t=2.00s,A = 0.15m, k= 10.0, and w(2.00s)=15.0
To find the wavelength,we will use the expression forwave number
K =
2𝜋
λ
rad/m
Where λ =
2𝜋
k
=
2𝜋
10.0
=
𝜋
5
= 0.628 m
To find the frequency,we lookat the given information and notice that we
can determine w first given w(2.00s)=15.0, and then determine the
frequency.
w(2.00s)=15.0
w= 7.50
f=
𝑤
2𝜋
=
7.50
2𝜋
= 1.19 Hz
Lastly, to find the velocity, we use the fact that
kv = w, as shown in the notes, and so v =
𝑤
𝑘
=
7.50
10.0
= 0.750 m/s
b) Since we are given t= 2.00, we can rewrite the equation as:
D(x, t) = 0.15m sin (10.0x-7.50t)
Problem 2:
A guitarstring generates a harmonicwave and it is described by the wave
function:
D(x, t) = 0.15m sin (2.5x-4.0t)

6. a) Find the velocity at t=0.0 s of a segment of the string located at x= 0.50
m
b) What is the maximum positivevelocityof this segment? When is the
first time after t=0.0 that the segment attains this velocity?
Solutions:
a) In the question,we are given a function of the displacement ofthe
harmonicwave. Prior to pluggingin the given informationto solvefor
an answer,we must take the derivativeof the displacement function
to arriveat an expression forvelocity
D(x, t) = 0.15m sin (2.5x-4.0t)
So
𝑑
𝑑𝑡
[0.15m sin (2.5x-4.0t)] = -0.60
𝑚
𝑠
cos (2.5x-4.0t)
Now we can plug in t=0.0s and x=0.50m
v(0.50m, 0.0s) = (-0.6
𝑚
𝑠
)[cos(2.5*0.50m) - 0s) = -0.19 m/s
b) The maximum velocityoccurs when the cosine function is equal to -1
since it gives a positivemaximum value of +0.60 m/s.
We simplyequate the velocityfunction to 0.60m/s and isolate for time t.
0.60 = -0.60
𝑚
𝑠
cos (1.25-4.0t)
-1 = cos (1.25-4.0t) *Cos(x)= -1 when x=±π
±π = 1.25-4.0t
t =
1.25±𝜋
4.0
(Note: 1.25 –π gives a negative time value, hence it is omitted)
so, t =
1.2+𝜋
4.0
= 1.09 s