phys 20 lessons unit 6: simple harmonic motion mechanical waves lesson 9: physics of music resonance...
TRANSCRIPT
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