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 half-

integer 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 half-

integer 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 trough-

to-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 + cosθ

2 = 2cos((θ

1-θ

2)/2)*cos((θ

1+θ

2)/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.