PHYS 20 LESSONS

Unit 6: Simple Harmonic

Motion

Mechanical Waves

Lesson 9: Physics of Music

Resonance and Standing Waves

Reading Segment #1:

Physics of Music

Resonance

To prepare for this section, please read:

Unit 6: p.27

Characteristics of Sound

Pitch

- the frequency of the sound

- humans can hear between 20 Hz and 20 kHz

- if the frequency is greater than 20 kHz, we call it

ultrasonic

- ultrasonic waves are detected by dogs, bats, and dolphins

- we use ultrasonic waves in technologies

such as ultrasound and sonar

Loudness

- the amplitude of the sound

- measured in decibels (dB)

0 dB is the threshold of human hearing

120 dB is a rock concert, and is damaging to

our ears

Quality

- the type or nature of the sound wave

- a pleasant sound has a regular (repeated) wave

e.g. Tuning Fork Violin

(pure tone) (overtones)

- irregular waves are called noise

Resonance

Natural Frequency

- each object has one (or more) natural frequency at which

it will vibrate

- you can easily determine the natural frequency of an object:

strike the object so that it begins to vibrate

the frequency at which it vibrates is a natural frequency

- detected by a microphone and studied using

an oscilloscope

e.g.

When tuning forks are struck, they have only one natural

frequency and they vibrate with a pure tone.

Examples of tuning forks would be:

256 Hz , 512 Hz , 1024 Hz , etc.

Tuning forks are useful for experiments, as well

as tuning musical instruments.

Resonance

- when a periodic force is applied to an object with the same

frequency as its natural frequency, the amplitude will increase

i.e. when the applied force and the natural frequency of

the object are in phase, the amplitude will increase

- this response (this increase in amplitude) is called resonance

Examples of Resonance:

Pushing a person on a swing

- you push in phase (or in time) with the natural

frequency of the swing

- if you do, the person swings higher and higher

- this increase in the swing amplitude is called

resonance

Video

Note:

Resonance must be taken into account in the design of bridges, propellers, turbines, car motors, etc.

If the frequency of the propeller's motion (for example)

matches the natural frequency of the motor materials, it causes

the motor to shake significantly.

The shaking can damage the motor.

Engineers use many materials and friction in the joints to

reduce this problem.

Video

To see the danger of resonance,

click on the following link:

Tacoma Narrows Bridge

Example

A car is stuck in a rut, and when the gas pedal is applied, the wheels simply spin out.

Using the principles of resonance, explain how the car can get out of the rut.

Try this example on your own first.Then, check out the solution.

Rocking a car to get it out of a rut

- you push at the same time that the driver pushes on

the gas pedal

- then, you let go when the car rocks back

- when it moves forward again, you push and the

driver uses the gas pedal again

- in this way, you are pushing in phase with the rocking

of the car

- the rocking of the car will increase (resonate) until

the car gets out of rut

Reading Segment #2:

Standing Waves

To prepare for this section, please read:

Unit 6: p.28

Standing Waves

When identical waves trains move toward each other and

interfere, a wave pattern emerges.

The resultant wave appears not to move (i.e. to stand in place),

resulting in regions of very high amplitude.

When this happens, we say that the material is resonating.

Animation

Standing Waves

1. http://www2.biglobe.ne.jp/~norimari/science/JavaEd/e-wave4.html

2. http://

members.aol.com/nicholashl/waves/stationarywaves.html

Animation

Standing Waves

3. http://surendranath.tripod.com/Applets.html

For this applet, go to:

"Waves"

"Transverse Waves"

"Adding Transverse Waves"

Select "continuous waves, equal amplitude".

Anatomy of a Standing Wave:

FixedFixed

EndEnd

Fixed Fixed

End End

Nodes

(N)

These are regions of constant destructive interference

(zero amplitude)

FixedFixed

EndEnd

Nodes

(N)

Notice that there is always a node at a fixed end.

Fixed Fixed

End End

Antinodes (A)

These are regions of periodic constructive interference

(maximum amplitude)

This is the resonance of the medium.

Node to node

0.5

Fixed Fixed

End End

N N

One full wavelength is shown in green.

Thus, the distance from node to node is half of a wavelength.

i.e.

The length of one loop is 0.5

Stringed Instruments

While many objects have only one natural frequency,

stretched strings have many.

When a string is plucked or struck, many frequencies

are created in the string.

The frequencies that match the natural frequency of the string

create standing waves and last.

(other frequencies interfere randomly and die out)

These standing waves resonate, creating loud

(i.e. large amplitude) sound.

Their sound is often amplified by a box (e.g. a guitar)

or a surface (e.g. a piano).

Note:

The natural frequencies of a string depend on:

- length

- mass (thickness)

- tension

We will focus specifically on the first factor: length.

Natural Frequencies of Stretched Strings

The natural frequencies of a string depend on:

- length

- mass (thickness)

- tension

We will focus specifically on the first factor: length.

Consider a string of length L, stretched between two fixed ends:

L

Fixed Fixed

End string End

We will now investigate the natural frequencies of this

string

i.e. the frequencies that would cause this string to resonate

These are called the harmonics of the string.

Fundamental frequency (n = 1): L

This is the lowest possible

frequency (f0 or f1).

At this frequency, one loop N A N

is created. This is the simplest

possible standing wave.

Remember, there must be a node at a fixed end.

L

At f = f0,

L = 0.5

N N

0.5 Use this formula only Node to node

when you are dealing with

the fundamental frequency

(called the first harmonic: n = 1)

Second Harmonic (n = 2): L

This is the next possible

frequency (f2).

At this frequency, two loops N N N

are created.

L

N N N

Notice that the wave has been shortened by 1/2.

What does that mean about the frequency?

From the universal wave equation f = v

we know that f 1

Frequency has an inverse relationship with wavelength.

This is assuming that the speed remains constant.

Since the medium is staying the same (e.g. same string),

the speed will remain constant.

f 1

So, if the wavelength has been multiplied by 0.5,

then the frequency has been divided by 0.5

i.e. The frequency has doubled

L

N N N

So, at the second harmonic: (n = 2)

f2 = 2 f0

The frequency has doubled

(compared to the fundamental frequency)

L

N N N

So, at the second harmonic: (n = 2)

f2 = 2 f0

L = 2 (0.5 )

Since two loops are created

General Formula:

For the nth harmonic: (n loops)

fn = n f0

L = n (0.5 )

v = fn n

When you use the universal wave equation, make sure

the frequency and wavelength are from the same harmonic.

Animation

Standing Wave in a Stringed Instrument

1. http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html

2.

http://www.physicsclassroom.com/Class/sound/U11L4c.html

Ex. Standing waves are created in a string by a source

vibrating at 120 Hz. Five loops are counted in a length

of 63.0 cm.

Find:

a) the wavelength

b) the fundamental frequency of the string

c) the speed of the wave

Sketch the standing wave first:

L = 0.630 m

This string is at the 5th harmonic (n = 5).

a) L = 0.630 m

L = n (0.5 )

= L = 0.630 m

0.5 n 0.5 (5)

= 0.252 m

2/5 of the length L

(2 loops out of 5)

b) At the 5th harmonic, the frequency is 5 times the

fundamental frequency.

So, fn = n f0

f0 = fn = f5 = 120 Hz

n 5 5

= 24.0 Hz

c) Based on the universal wave equation

v = f5 5 Use the frequency and

wavelength from the same

= (120 Hz) (0.252 m) harmonic

= 30.2 m/s

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 6 p. 29 #1 - 6

Ex. A 55 cm long string resonates at the 4th harmonic

when the source frequency is 380 Hz.

This same string is then cut to a length of 37 cm

(tension remaining the same). What is its

fundamental frequency?

Key Strategy:

Since the string has the same tension in both cases,

the medium stays the same.

This means, the waves through the string will have the

same speed.

So, find the speed in String 1, then use this speed for string 2.

String 1: L = 0.55 m

L = n (0.5 )

= L

0.5 n n = 4

= 0.55 m = 0.275 m

(0.5) (4)

The wavelength is half of

the total length

i.e. 2 loops out of 4

L = n (0.5 )

= L

0.5 n

= 0.55 m = 0.275 m

(0.5) (4)

Then, v = f

= (380 Hz) (0.275 m)

= 104.5 m/s It will be the same speed for

String 2.

String 2: L = 0.37 m

L = n (0.5 )

= L

0.5 n n = 1

= 0.37 m = 0.74 m

(0.5) (1)

L = n (0.5 )

= L

0.5 n

= 0.37 m = 0.74 m

(0.5) (1)

Then, v = f

f0 = v = 104.5 m/s = 141 Hz

0.74 m

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 6 p. 29 #7, 8