SoundSoundProperties

Sound Properties

• Sound Wave

Sound Properties: Detection

• Microphone– Convert kinetic energy to electricity

• Ear– Measure in decibel (dB)

Sound Properties: Frequency

• Pitch vs. Frequency– pitch is relative (a matter of common agreement among

musicians), while frequency is absolute (a precise, unambiguous measurement).

Sound Properties: Loud

• Amplitude is loudness– Read decibel

article

Sound Properties: Doppler Effect• Perceived frequency

– How often the pressure front hits your ear per second

• Scenarios– Sound source stationary, you stationary– Sound source moving, you stationary– Sound source stationary, you moving– Sound source moving, you moving

Sound Properties: Doppler Effect

Doppler effect equations

frequency perceived by detector

frequency of the wave

velocity of the wave

velocity of the detector

velocity of the source

w dd w

w s

d

w

w

d

s

v vf f

v v

f

f

v

v

v

Sound and Music

Sources

• Resonators• Strings• Sound boards/surfaces• Emmet Otter

Resonators

• Resonance– Not just reflection– Increases the amplitude of a vibration by repeatedly

applying a small external force at the same natural frequency.

– Closed Pipe (pg 413, fig 15-11)– Open Pipe (pg 413, fig 15-11)– Displacement and Pressure

Resonance

• Closed Pipe: 1 2 3

4 44 , ,

3 5L L L

Resonance

• Open Pipe 1 2 3

22 , ,

3L L L

Resonance

• Strings L

2L

L

2

3L

Consonance and Dissonance

• Consonance– Combination of pitches that is pleasant

• Dissonance– Combination of pitches that is not pleasant

Beat

• Two frequencies are very close, causes oscillations in amplitude

Binaural Beat Brain Wave

ReviewReviewDoppler Effect

Doppler Effect

• The source of the sound is moving toward the detector

• The detector is moving toward the sound source• The source is moving away from the detector• The detector is moving away from the source• The source and detector are moving toward each

other• The source and detector are moving away from

each other– Which situations result in an increased frequency and

which result in a decreased frequency

Doppler Effect

frequency perceived by detector

frequency of the wave, often called the source

velocity of the wave, often called the source

velocity of the detector

velocity of the so

w dd w

w s

d

w

w

d

s

v vf f

v v

f

f

v

v

v

urce

Decibels

• 3dB increase is double power• 3dB decrease is half power• 20dB increase is 10x as much power• 20dB decrease is 1/10 as much power

Pipe resonator

• Closed pipe– ¼ wavelength at fundamental harmonic

• Open pipe– ½ wavelength at fundamental harmonic

• Each successive harmonic is ½ wavelength higher

or• The distance between any two consecutive

harmonics is ½ wavelength

Assumptions

• Assume that the speed of sound in air is 343 m/s, at 20°C, unless otherwise noted.

• Assume that the speed of sound in water is 1533 m/s, at 25°C, unless otherwise noted.

• Assume that the speed of sound in water is 1530 m/s, at 20°C, unless otherwise noted.

Problems

18.5Hz

Problems

• An open-pipe resonator has a length of 2.39m. Calculate the frequency of its third harmonic if the velocity of sound is 343 m/s.

Problems

• You are listening to an outdoor concert on a day when the temperature is 0°C. The sound of a wavelength of 0.490 m is emitted by a flute on the stage 125 m from where you are standing.– a.What is the time elapsed before you hear the sound emitted from

the stage?

– b.What is the frequency of the sound?

Problems

• The pulse-echo technique is used in diagnostic medical imaging. A short ultrasound pulse is emitted from the device, and echoes are produced when the pulse is reflected at a tissue interface. The echo signals are received back at the device and then analyzed to build up an image of the organ. The speed of sound in soft tissue is 1540 m/s. If an echo is received 58.210–6 s after the pulse was emitted, how far is the tissue interface from the ultrasound device?

Problems

• The engine of a jet plane taking off produces a sound level of 140 dB, and the sound wave has a pressure amplitude of 200 Pa. A baggage handler working next to a jet plane that is taking off is wearing specially designed hearing protectors that reduce the sound level entering his ear by 40 dB. What is the pressure amplitude of the sound waves entering his ear?

Problems

• While fishing from a boat anchored offshore, you see another fishing boat between your boat and the shore. The other boat sounds a 510-Hz horn as it heads toward the shore at a speed of 18 m/s.– a. If your fishing boat is stationary, what is the frequency of the

sound waves from the horn that reach you?

– b. If your fishing boat now heads out to sea at a speed of 15 m/s, what is the frequency of the sound waves from the horn that reach you?

Problems

• A species of bat navigates by emitting short bursts of sound waves that have a frequency range that peaks at 58.0 kHz.– a. If a bat is flying at 4.0 m/s toward a stationary object, what is

the frequency of the sound waves reaching the object?

– b. What is the frequency of the reflected sound waves detected by the bat?

– c. What is the difference between the frequency of the sound waves emitted by the bat and the frequency of the sound waves detected by the bat if the bat is flying at 4.0 m/s and the object is a moth approaching at 1.0 m/s?

Problems

• Hannah places an open, vertical glass tube into a container of water so that the lower end of the tube is submerged. She holds a vibrating tuning fork over the top of the tube while varying the water level in the tube. Hannah notices that the loudest sound is heard when the distance from the water to the top of the tube is 32.7 cm, and again when the distance is 98.2 cm. What is the frequency of the tuning fork?

Problems

• The six strings of a standard guitar are tuned to the following frequencies: 165, 220, 294, 392, 494, and 659 Hz.– a. Find the lengths of the shortest open-ended organ pipes that

would produce the same frequencies.

– b. Sketch the pipes, showing their lengths to scale.

Problems

• The fundamental tone of an open-pipe resonator with a length of 48 cm is the same as the second harmonic tone of a closed-pipe resonator. What is the length of the closed-pipe resonator?

Problems

• You receive a CD with the following note: “The first sound on the CD is the sound of a 238-Hz tuning fork and a second tuning fork being struck simultaneously. The second sound on the CD is the sound of the second tuning fork and a 240.0-Hz tuning fork being struck simultaneously. What is the frequency of the second tuning fork?” Listening to the CD, you hear that the first sound has a beat frequency of 3.00 Hz and the second sound has a beat frequency of 5.00 Hz. Answer the question found in the note.

• fbeat = f2 – f1

• (f2 – f1) = fbeat

• f2 = f1 fbeat

• = 238.0 Hz 3.00 Hz

• = 241 Hz or 235 Hz

• fbeat = f2 – f3

• (f2 – f3) = fbeat

• f2 = f3 fbeat

• = 240.0 Hz 5.00 Hz

• = 245 Hz or 235 Hz

• The frequency of the second tuning fork must be 235 Hz.

Problems

• A radio station broadcasts their signal with a wavelength of 3.5 µm. Although your radio will translate this signal into audible sound, explain why you cannot hear the radio signal directly.

f v

343 m/s

3.5 10 6 m9.8 107s 1

98MHz

The threshold of the human ear is around 20,000 Hz, so the frequency of this radio signal is far higher than what the ear can detect.

NO

Problems

• A baseball fan sits in the outfield seats watching his home team play while another fan watches the same game at her house on television. In his seat at the ballpark, the fan sits 134 m from home plate. At her house, the other fan sits 2.0 m from the television speaker, watching a signal broadcast from a camera located 8.0 m behind home plate. Assume the temperature throughout the city is 30.0°C and that there is no time delay in the television transmission. The TV signal travels at c.– a. The batter hits a fly ball. Which fan hears the crack of the bat

first? Why?– b. A third fan hears the crack of the bat a full 2.00 s after she sees

it. How far away is she?

a. The sound of the crack of the bat travels at a speed of

v = 331 m/s + 0.6T

= 331 m/s + (0.6)(30.0°C)

= 349 m/s

For the fan sitting in the seats at the ballpark,

v d

t

t d

v

134 m

349 m/s0.384 s

• For the fan sitting at home, the time of video transmission of the sound is negligible, as radio signals travel at the speed of light, c = 3.0108 m/s. Even if the fan is watching from 1000 km away, the time of travel for the video signal is only 3.3 ms.

• The fan watching the game on television actually hears the crack of the bat before the fan in attendance at the ballpark.

d = 8.0 m + 2.0 m = 10.0 m

t d

v

10.0 m

349 m/s0.0287 s

• b.

v d

t

d vt

349 m/s 2.00 s

698 m

Problems

• An engineer at an underwater military station listens for submarines by sending an ultrasound sonar ping that has a frequency of 3.75 MHz.

• a. A stationary object is detected when the ping returns 3.00 s later. How far away is this object? The speed of sound in seawater is 1533 m/s.

• b. A second ping returns with a frequency of 3.80 MHz, indicating that the object is now moving. What is the object’s velocity? In which direction is it moving relative to the listening station? Hint: The direction of sound reverses after the sound reflects off the moving object.

• c. Sonar equipment has difficulty detecting objects smaller than the wavelength of the ping. Old sonar equipment used an audible ping with a frequency of 4.00102 Hz. What is the smallest object this old sonar could distinguish?

• d. What is the smallest object the ultrasound sonar can detect?

• a. A stationary object is detected when the ping returns 3.00 s later. How far away is this object? The speed of sound in seawater is 1533 m/s.– The distance for the sound to travel and return is twice the

distance to the object: v 2d

t

d 1

2vt

1

21533 m/s 3.00 s

2.30 103 m

• b. A second ping returns with a frequency of 3.80 MHz, indicating that the object is now moving. What is the object’s velocity? In which direction is it moving relative to the listening station? Hint: The direction of sound reverses after the sound reflects off the moving object.

• where fs is the frequency of the sonar ping at the source and v is the velocity of sound in seawater. For the echo, the source (now the submarine) is moving, and the sound wave travels in the return direction. The frequency that the engineer detects is fd2, and the frequency at the source is fd1.

f d1 f s

v v sub

v

• Substitute for fd1, and solve for vsub.

f d2 f d1

v

v v sub

f d1v

v v sub f d2 f s v v sub

v v sub

v sub v f d2 f s

f d2 f s

1533 m/s 3.80 MHz 3.75MHz

3.80 MHz 3.75MHz

10.2 m/s

Since the frequency of the sonar ping increased, the ship must be approaching the engineer.

• c.

• d.

v

f

1533 m/s

4.00 102 Hz3.83 m

v

f

1533 m/s

3.75 106Hz

4.09 10 4m

=0.409 mm

Problems

frequency perceived by detector

frequency of the wave

velocity of the wave

velocity of the detector

velocity of the source

w dd w

w s

d

w

w

d

s

v vf f

v v

f

f

v

v

v

Elements

More abstract animations like these can be found on the Premium Gold Site. Search Bloodhound for “Strange Mechanical”.