Academic Resource Center:
Waves Workshop
Presentation Outline
•Understanding concepts
Types of waves
• Transverse Waves
• Longitudinal Waves
Interference
• Constructive v. Destructive
• Beating
•Practice problems
Singing in the shower
Tuning instruments by ear
Caterpillar motion
Interference
Wave Types
Transverse Waves: Disturbance perpendicular to propagation
Ocean waves
caterpillar
Longitudinal Waves: Disturbance parallel to propagation
Traffic jams
sound
Interference
Constructive interference
Destructive interference
Interference
When two waves are similar in frequency or phase, but not exactly
matching, they “beat” together.
Practice Problem: Singing in the Shower
Many men like singing in the shower stall because somehow enhances their
voice. How does this happen? Would the effect be different for men and
women?*
*Problem 17.8 Ohanian Physics for Engineers 3rd Editions
Shower Singing: Solution
Sound waves are trapped within the shower stall and bounce back and forth
from the singer to the wall, as well as wall to wall. The sound waves interfere
with themselves this way.
If the distance between the wall and the singer (or wall-to-wall) is an halfinteger multiple of the wavelength, constructive interference creates standing
waves which amplify the sound.
Here the distance from singer to wall is 1.5*wavelength of the sound. The
sound wave is amplified.
Shower Singing: Solution
If the distance between the wall and the singer (or wall-to-wall) is not an halfinteger multiple of the wavelength, destructive interference occurs.
Because women's voices are typically slightly higher than men's voices, the
wavelengths of their singing would be slightly shorter than a half-integer
multiple. This causes a beating phenomenon which sounds off-key.
Practice Problem: Tuning
String instruments such as the guitar or piano need to be tuned often; the
tension on the strings need to be adjusted so that each string plays a certain
musical note.
Expert players can tune strings by ear as they have before modern technology.
One method involves playing an instrument that is already tuned and adjusting
your instrument accordingly.
Say you have one guitar that is properly tuned and another that is not. How
can you tell when a particular string is tuned just right? (Hint: what would
happen if the string was only slightly off-tune? What would it sound like?)
Tuning: Solution
When you tune a string close to the right tension, the sound it would produce
would have a wavelength close, but not exactly, to wavelength it should have.
For example the low E string should produce a sound with wavelength 1670cm,
but when it is slightly out of tune it produces 1650cm wavelength sounds.
When the tuned string and out-of-tune string are plucked at the same time, the
two waves, that have similar wavelengths, beat together. The sound of the
beating would oscillate between loud and quiet until they both die down.
Practice Problem: Caterpillar
A nature photographer took this picture of a caterpillar moving on a leaf. After
taking the picture, he observed it to move at about 1 inch every 2 seconds.
Estimate the wave number and frequency of the wave traveling through the
caterpillar's body as it moves.
Caterpillar Problem: Solution
The first quantity to be solved is wave number. The definition of wave number
is 2π/λ.
What is λ? It is the wavelength of the wave traveling through the caterpillar.
Any wavelength can be measured from crest-to-crest,node-to-node, or troughto-trough.
The picture can be used. From one end of the caterpillar to the other measures
about 1 inch using the scale, approximately.
Therefore 1” = λ;
k = 6.3 in-1
Caterpillar Problem: Solution
The second quantity to be solved is frequency. There are many relations
involving frequency, e.g. ω = 2πf, however always look for the relationship that
has what you are solving for (f) and what you already know (λ, k, v).
You should notice you did not use one piece of information yet, the speed (1
inch per 2 seconds). This is a hint to find the relation between f and v.
λf = v, or f = v/λ
f = (1”/2sec)/(1”) = 0.5sec-1 = 0.5 Hz.
Practice Problem: Interference
Two transverse harmonic waves are described by
y1 = Acos(πx - 3πt)
and
y2 = Acos(πx + 3πt)
Where A = 5.0m, x is in meters, and t is in seconds. What is the maximum
amplitude of the superposition of these two waves at x = 0.25m? What are the
maximum transverse speed and acceleration at that point?*
*Problem 16.59 Ohanian Physics for Engineers 3rd Editions
Interference Problem: Solution
Setting x = 0.25cm, the super position of both waves can be described as
y = 5cos(π/4 – 3πt) + 5cos(π/4 + 3πt)
The two cosines can be combined using the identity
cosθ1 + cosθ2 = 2cos((θ1-θ2)/2)*cos((θ1+θ2)/2)
Therefore
y = 5*2*cos(π/4+3πt)*cos(0) = 10cos(π/4 + 3πt)
The maximum value of cosine is when the argument is zero. When it is zero,
y = 10*cos(0) = 10
The maximum amplitude possible is 10m.
Interference Problem: Solution
Speed and acceleration can be obtained by taking the first and second
derivatives of the position y.
y = 10cos(π/4 + 3πt)
v = dy/dt = -30πsin(π/4 + 3πt)
a = d2y/dt2 = -90π2cos(π/4 + 3πt)
The maximum speed is 30π m/s while the maximum acceleration is 90π2 m/s2.
Note that these are maximum values; the actual speed and acceleration vary
with time at the position x = 0.25m.