Harmonic waves

1. Harmonic Waves
Wave – defined as the motion of disturbance
Examples:
Mechanical wave Electromagnetic wave:
- travels through a medium and
wave speed depends on the
medium


- generated by charged particles
and travels independently of
the medium






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When a source creates a wave that travels in simple harmonic motion, it is
referred as a harmonic wave or sinusoidal wave.
Properties of Harmonic Waves
Amplitude: maximum displacement of the medium from equilibrium
position (±A)
The maximum positive displacement (+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 by lambda, λ.

2. http://nothingnerdy.wikispaces.com/4.4+WAVE+CHARACTERISTICS

The vertical displacement of an element at location x can be calculated using
the following general equation:
D(x) = A sin(kx)
Notice the new quantity wave number (k) in the given equation. This
quantity has no correlation with the spring constant k! Rather, it represents
the how often a wave pattern repeats per unit distance in this context.
K =
2??????
λ
rad/m
Period (T): the time it takes complete one wave cycle
Frequency (f): the number of wave cycles passing a 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 passing a fixed point of a medium, each
with a length λ. The wave speed is therefore: V= ???????????? and this holds true for
any kind of periodic waves.
Previously, we introduced wave number, k, and so ?????? can be rearranged as
2??????
k
.

3. The wave speed can be rewritten in terms of wave number and angular
frequency: v= (
2??????
k
)(
??????
2??????
) =
??????
??????

As time changes, the displacement of any element of a medium changes
from its equilibrium position. Therefore, the function for a harmonic wave
can be expressed differently depending on whether it is travelling in the
direction of increasing x or decreasing x.
Increasing x: 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 (
????????????
??????
x -
????????????
??????
t)
*Note: kv = w, so kvt = wt
Decreasing x: D(x,t) = A sin(
????????????
??????
x +
????????????
??????
t)
However, notice how in the above functions when the position is at x=0 and
t=0, the displacement is restricted to being zero only.
Here is when the phase constant is added to the function to account for this
restriction. The functions can now be rewritten in terms of wave length and
period as well as the phase constant ϕ:
Increasing x: D(x,t) = A sin (
????????????
??????
x -
????????????
??????
t + ϕ )
Decreasing x: D(x,t) = A sin(
????????????
??????
x +
????????????
??????
t + ϕ)

4. Transverse Velocity and Acceleration
Each segment of a medium that has a harmonic wave travelling through it
experiences a different velocity as time changes
The velocity is the rate of change of the displacement of the 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 changing its velocity.
Similarly, the rate of change of the instantaneous velocity of the segment
gives acceleration
a(x,t) = -w
2
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
itself has no acceleration.
Step by step solutions to end of chapter problems:
Problem 1:
The displacement function at t=2.00 s for a travelling wave is given by the
following equation:
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 for D(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 for wave number
K =
2??????
λ
rad/m
Where λ =
2??????
k
=
2??????
10.0
=
??????
5
= 0.628 m
To find the frequency, we look at 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=
??????
2??????
=
7.50
2??????
= 1.19 Hz
Lastly, to find the velocity, we use the fact that
kv = w, as shown in the notes, and so v =
??????
??????
=
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 guitar string generates a harmonic wave 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 positive velocity of 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 of the
harmonic wave. Prior to plugging in the given information to solve for
an answer, we must take the derivative of the displacement function
to arrive at an expression for velocity

D(x, t) = 0.15m sin (2.5x-4.0t)

So
??????
??????�
[0.15m sin (2.5x-4.0t)] = -0.60
??????
�
cos (2.5x-4.0t)
Now we can plug in t=0.0s and x=0.50m
v(0.50m, 0.0s) = (-0.6
??????
�
)[cos(2.5*0.50m) - 0s) = -0.19 m/s

b) The maximum velocity occurs when the cosine function is equal to -1
since it gives a positive maximum value of +0.60 m/s.
We simply equate the velocity function to 0.60m/s and isolate for time t.
0.60 = -0.60
??????
�
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±??????
4.0
(Note: 1.25 –π gives a negative time value, hence it is omitted)
so, t =
1.2+??????
4.0
= 1.09 s