SPH 3UO LESSON PLANS


Unit 3: Waves and Sound
- Mechanics of waves
- Properties of waves
- Sound as a wave
- Resonance
- Music
- Societal impact of waves and sound

Text: Physics 11
(Addison-Wesley)
 
 

Lesson One
Summary:
-view Bill Nye video
-vibrations(definition, types, frequency, period)
Practice:
-do p. 239 #1,2
Review:

-do p. 
240 #1-7

VIBRATIONS

Periodic Motion - motion that repeats itself...each repeated motion is called a cycle.
Types:
1)Transverse - if the object moves perpendicular to its length
ex. water saves, where A is the amplitude and w is the wavelength (sorry, I can't do that symbol)

wavelength (w) = distance between two points vibrating "in phase"...distance from crest to crest or trough to trough
amplitude (A) = maximum distance from the zero point...distance from the zero point to each trough or crest

2) Longitudinal -if the motion of the object is parallel to it's length
ex. .       .      .      .... .        .        .        .       .       .... .    .
                    compression      rarefaction            compression
wavelength goes from compression to compression
T (period): the time for one complete cycle
T= /\t/N       where /\t = time interval in s, N = #of cycles (Hz)
f (frequency): the number of complete cycles in certain time
f = N//\t




Lesson Two
Summary:
-mechanical resonance
-view resonance and waves video
-types of waves
-wave machine demo

-do
discovery, p.237
Review:
-do p. 243 #1-6
-do p. 246 #1-4
Assignment:
-do
discovery, p.237

MECHANICAL RESONANCE


Lesson Three
Summary:
-speed of waves
-doppler effect
-wave behaviour
(interference,diffraction,
refraction)
-do
discovery, p.249
-do discovery, p.253
-do discovery, p.256

Practice:
-do p. 248 #1,2
-do p.
256 #1,2
Review
:

-do p. 251 #1-7,9
-do p. 257 #1-8
-do p. 262, #15,20

Assignment:
-do discovery, p.249
-do discovery, p.253
-do discovery, p.256

Due:

-quiz on waves next class

WAVE BEHAVIOUR

The Universal Wave Equation
v= fw...where v = speed (m/s), f = frequency (Hz), w =wavelength (m)
just like d=vt from before

Interference - were particles meet there is a collision, but waves are different
1) constructive interference
-amplitudes add to make a bigger wave
ex.a wave with amplitude  of 3 meets a wave with and amplitude of 5
 ...when they combine the wave has an amplitude of 8, but when the two waves move on they return to their original amplitude

2) Destructive interference
-amplitude add to make a smaller waves
ex. a wave with a negative amplitude of 3 and wave with a positive amplitude of 5 meet
  ...when they collide they result in a positive amplitude of 2, and then both waves return to their original amplitude after they pass

3) Complete destructive Interference...both amplitudes are equal and add to zero
The point of collision that remains at rest the whole time is called the nodal point or node
 

Diffraction...waves tend to spread out after passing through an opening
              ...This effect is maximized if w = d where d is the size of the opening. 
                 If w<<<d then there is not much diffraction.

Refraction...waves travel at different speeds in different mediums.
                  This can cause the wave to change wavelength and sometimes bend.
                  The frequency remains constant.

Example 1: A 17 cm sound wave is moving at 3.4 x 102 m/s.
                 Calculate the frequency and the period of vibration.

Example 2: A wave with a speed of 100 cm/s and a wavelength of 8 cm passes into a second medium in which its speed becomes 60 cm/s.
                 What will be the wavelength of the wave in the second medium?


Lesson Four
Summary:
-quiz on waves
-producing sound and characteristics
-do discovery, p.265
-sound case study

-case study presentations begin in three classes

Review:
-do p. 
270 #1-5
Assignment
:

-do discovery, p.265
-case study: effect of noise on body
Due:
-case study in three classes


SOUND

Speed of Sound... v = 331m/s + 0.59xT in the air (T= temperature in oC)
-if the T = 0 oC then vsound faster in warm weather
-speed increases as one moves from gases to liquids to solids (table 7.4, page 278)
Sonar - sound navigation and ranging
measure t (in s for sound to travel out and back), you know v in water at that temperature
use d = vt to find depth of water

Sound- the source of all sound is a vibrating object, and requires a medium to travel through
ex. tuning fork and speaker vibrate to produce longitudinal waves travelling through the air
-sound is the form of energy that can be detected by the human ear
-a healthy young human has an audible range of 16 Hz- 20 kHz

Pitch - depends on the frequency
-high frequency means high pitch
-all frequencies sound different


Lesson Five
Summary:
-hearing sound (intensity)
-the human ear (homework)

-speed of sound

-echoes

-sound barrier

-do
discovery, p. 280

Practice:
-do p. 277 #1,2
-do p. 
279 #1-3
Review:

-do p. 276 #1-11
-do p. 280 # 1-7
Assignment:
-do discovery, p. 280

Due:

-case study in two classes


SOUND INTENSITY

Intensity (loudness)
-the standard unit for intensity is W/m2 but the practical unit is decibel (dB)
-threshold of hearing 0 dB
-threshold of pain 120 dB
These have a difference of 1 000 000 000 000 x
ex. a sound of 50 dB is 10x louder than a 40 dB sound
ex. a sound of 70 dB is 100x louder than a 50 dB sound

ex. speakers...there are nodal lines where the sound waves cancel and there is no sound

Beat frequency...a phenomena that occurs when you hear two notes together that are very close in frequency
You hear (f1 + f2)/2 (the average) but the intensity wobbles with a beat frequency of f1 + f2
ex. f1= 500 Hz , f2= 498 Hz....what do you hear? 499 Hz with a beat frequency of 2 Hz

The Doppler Effect...when a source of waves is moving the frequency depends on where the observer is...
source at rest = all waves same frequency
source moving to the right, high frequency to the right, low frequency to the left
What if you are moving to the right at the speed of sound?
High pressure will occur on the right as the sound barrier is a wall of air.



Lesson Six
Summary:
-sound test
-interference and diffraction
-extra time to work on case study

-do
discovery, sound test

Practice:
-do p. 282 #1,2
Review:

-do p. 284 #1-9
-do p. 288,#13,19,24,25,27
Assignment:
-do discovery, sound test

Due:

-case study next class

-sound quiz next day


INTERFERENCE

Interference: when two sound waves of the same f meet thay can add or subtract
                  in phase they would result in a loud noise
                  out of phase they would result in no sound

Resonance: a small force at the right frequency can cause large vibrations.
                   These frequencies are called resonant frequencies.

                   ex. tuning forks, bridges, eyelids...small force can make them vibrate


Lesson Seven
Summary:
-sound quiz
-case study presentations
-distribute instrument project

Due:
-instrument project in four classes

CASE STUDY PRESENTATIONS


Lesson Eight
Summary:
-start instrument project
-standing waves
-reflection

-do discovery, p. 
291
-do
discovery, p. 293

-do
discovery, p. 296

Review:
-do p. 296 #1-6
Assignment
:

-do discovery, p. 
291
-do
discovery, p. 293

-do
discovery, p. 296

Due
:

-instrument project in three classes

STANDING WAVES

Standing Waves: If you have a rope that is fixed at both ends, L = length of the rope
f1  for one loop, f1 = V/w1 , w1 = 2L     ==> f1 = V/2L
f2  for two loops, f2 = V/w2 , w2 = L     ==> f2 = V/L = 2f1
f3  for three loops, f3 = V/w3 , w3 = 3L/2     ==> f3 = 3V/2L = 3f1
f8  for eight loops, f8 = 8f1
There are the resonant frequencies of the string
The first is called the fundamental frequency
All other are multiples of the fundamental frequency
At any other frequecy you do not get standing waves
When a guitar string is struck it vibrates at the fundamental frequency
f1 = V/2L
If you want that string to play different f then you can change V or L. L is easier.
ex. A string 20 cm long plays at 392 Hz. How long should it be to produce a 523 Hz note?
f1 = V/2L find V!
v= f1 2L
v = (392)(2)(0.2) = 156.8 m/s, v doesn't change
f1 = V/2L find L!
523 = 156.8/2L
L = 15 cm

 


Lesson Nine
Summary:
-stringed instruments
-do musical instrument project

-do
discovery, standing wave demo

Practice:
-do p. 298 #1,2
Review:

-do p. 300 #1-8
Assignment:
-do discovery, standing wave demo 
Due:

-instrument project in two classes


STRINGED INSTRUMENTS




Lesson Ten
Summary:
-air columns
-wind instruments
Practice:
-do p. 302 #1,2
-do p.
303 #1,2
-do p. 
304 #1,2
-do p. 310 #1,2
Review:

-do p. 306 #1-7
-do p. 313 #1-9
Assignment:
-do discovery, air column demo
Due:
-instrument project next class

AIR COLUMNS

Air Columns...
recall: for a string fixed at both ends each end has to be a node
In an air column with one open end and one closed end, sound of a certain frequency can resonate
open end must by an anti-node, closed end must be a node
for 0.5 loops, L = 1/4 w
for 1.5 loops, L = 3/4 w
for 2.5 loops, L = 5/4 w

ex. To resonate a sound of 1024 Hz what lengths of air column will work?
The lengths that will work are w/4, 3w/4, 5w/4,...
v = fw
w = v/f

to find w, we need v
v= 331 + 0.6T
v= ...

Resonance box:
find f of the tuning fork...
L= ___
w = 4L
f = v/w
f =...
 

For an air column that is open at both ends, L = 1/2 w, and an anti-node is at both ends
Make a straw that will amplify a tuning fork

Quality of Sounds
If you hear the note A (440 Hz) with equal intensity from a piano and trumpet they both sound different
The note A
f1 = 440 Hz + first harmonic
f2 = 880 Hz + second harmonic
f3 = 1320 Hz + third harmonic
etc....
Any string or air column that resonates 440 Hz will also resonate 880 Hz, 1320 Hz
Ex. You play A on instrument 1 and instrument 2
Instrument 1 has 100% 440 Hz, 70% 880 Hz, 20% 1320 Hz
Instrument 2 has 100% 440 Hz, 60% 880 Hz, 10% 1320 Hz
, sound from instrument 1 adds to give a complicated wave that looks different from instrument 2
...They sound different only because they play the harmonics in different ratios.



Lesson Eleven
Summary:
-instrument project
Due:
-test next class

MUSICAL INSTRUMENTS
 


Lesson Twelve
Summary:
-test


SOUND TEST




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