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## PowerPoint Slideshow about 'Properties of Sound' - linh

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There are several ways in which sounds can differ from one another.

- they can be louder or softer

- they can be of a high or low pitch

- they can be melodic or just “noisy”

Measurable Quantity

effect on ear

Intensity

loudness

frequency

pitch

harmonic content

quality

The Intensity (I) of a sound source is directly related to the power (P) of the source and inversely related to the spherical area (A) over which it is spread.

P

A

Spherical:

A = 4r2

units:

W/m2

I =

Loudness is the effect of intensity on the ear:

- generally, more intense sounds seem louder

- it is subjective and not directly quantifiable

For any given frequency, the highest intensity at which the human ear can distinguish sounds is called the Threshold of Pain (great name for a band!)

- While this varies a bit from f to f and ear to ear, it is an average value of 10 W/m2

- Any more intense than this and only pain is experienced- can’t tell a dog barking from a piano playing.

For any given frequency, the lowest intensity that the human ear can pick up is called the Threshold of Hearing (I0).

- It varies from ear to ear and f to f, but it has an average value of 10-12 W/m2

Because of the great range of intensities the ear is sensitive to, a logarithmic comparative scale is used to discuss Relative Sound Intensity ().

Relative Sound Intensity is a logarithmic comparison of the intensity of a sound to the intensity of the Threshold of Hearing:

I0 = 10-12W/m2

The equation for Relative Sound Intensity: ear can pick up is called the Threshold of Hearing (I

units:

decibels (dB)

a dimensionless unit!

= 10log(I/I0)

The Threshold of Pain would be about 120-130 dB and the Threshold of Hearing would, of course, be 0 dB.

- close to a jet engine 150 dB

- general speech 40 to 80 dB

A small sound source emits sound energy at the rate of 2.00 W. What is the relative sound intensity a point 34.0 m away from the source?

P = 2.00 W

= 10log (I/I0)

d = 34.0 m

= 10log[(1.38 X10-4)/(10-12)]

= ?

= 10log(1.38 X 108)

I0 = 10-12W/m2

= 10(8.14)

I = P = 2.00 W

A 4(34.0m)2

81.4 dB

= 1.38 X 10-4 W/m2

A plane is found to have a relative sound intensity of 115 dB. What must be the intensity of the plane’s sound?

= 115 dB

= 10log(I/I0)

I = ?

115 = 10 log (I/10-12 W/m2)

11.5 = log (I/10-12 W/m2)

1011.5 = (I/10-12 W/m2)

I = (1011.5)(10-12 W/m2)

= .316 W/m2

What is the relative sound intensity at a point that is 2.00 m away from a 20.0 W sound speaker?

At a point that is 15.0 m away from a sound source the relative sound intensity is measured to be 102 dB. What is the power of that sound source?

A single monkey produces a relative sound intensity of 83.0 dB at a given point in the monkey exhibit at the zoo. What would be the relative sound intensity of all 15 monkeys chattering at once?

116 dB, 44.6 W, 94.7 dB

1) A sound source will produce a relative sound intensity of 81 dB at a point that is 34 m away from the source. If the power of the source were doubled, and the distance to the source were cut in half, what would the relative sound intensity be then?

2) A packaging machine in a factory is found to have a relative sound intensity of 79.5 dB when it is running full speed. What would be the relative sound intensity of 24 such machines running at full capacity?

At a rock concert, the relative sound intensity can be 125 dB. How far from the stage was this intensity measured if the sound is produced 825 w?

When 400 students are in the auditorium, the relative sound intensity is measured at 84.0 dB. What is the relative sound intensity if all 1350 students are gathered?

The relative intensity of a sound is measured at 115 dB at a distance of 3.50 m from a source. What is the power of the source?

Frequency and Pitch dB. How far from the stage was this intensity measured if the sound is produced 825 w?

Frequency (number of waves per second) produces the effect of pitch on the ear.

- A listener in front or behind or to the side of a source will all hear the same pitch as long as both listeners and source are stationary.

- If either the sound source or the listener is moving, a different pitch will be heard due to frequency shift.

- This shift in frequency from the first is due to the Doppler Effect.

A stationary sound source: dB. How far from the stage was this intensity measured if the sound is produced 825 w?

Same f and same in all directions!

If a sound source is moving: dB. How far from the stage was this intensity measured if the sound is produced 825 w?

The wavelengths in front get compressed and in back they get lengthened!

f is inversely related to since v is constant

f in front is higher and f behind is lower!

The faster the source travels, the more the wavelengths will be compressed:

When the source moves at the speed of sound no wavelengths exist in front of source- no sound!

Sonic Boom

Doppler Effect be compressed: A change in pitch due to the relative motion of a sound source, a listener, or both.

Doppler Examples

If the source is moving:

- A listener in front will hear a higher frequency than the source actually emits.

- A listener behind a moving source will hear a lower frequency than the source.

flf = fsv

v - vs

flb = fsv

v + vs

If the listener is moving relative to a stationary sound source, the frequency will again vary:

fL = fs( v ± vL)

v

If the listener is moving toward the source: +

If the listener is moving away from the source: −

If both the listener and the source are moving:

We will not consider this case during this class.

fL = fs( v ± vL)

(v ± vs)

A car approaches you at a speed of 24.0 m/s sounding a horn of frequency 456 Hz. If the air temperature is 23.0˚C, what frequency will you hear?

flf = fsv

v - vs

= (456 Hz)(345.3 m/s)

(345.3 - 24.0)m/s

vs = 24.0 m/s

fs = 456 Hz

v = 331.5 m/s + (.6)(23.0)

= 490 Hz

345.3 m/s

flf = ?

If this car is stationary and sounding its horn, what frequency would you hear if you drove towards it at 24.0 m/s?

As a train pulls away from a station, a listener on the platform hears a frequency of 285 Hz. However, the engineer hears a frequency of 322 Hz. If the air is 22.0˚C, what must be the speed of the train?

flb = fsv/(v + vs)

flb = 285 Hz

vs = fsv/flb - v

fs = 322 Hz

=(322 Hz)(344.7m/s) - 344.7m/s

285 Hz

v= 331.5m/s + (.6)(22.0) = 344.7 m/s

= 44.8 m/s

vs = ?

1) What is the drop in frequency heard as a train approaches and passes a listener if the train is traveling at 19.4 m/s and horn has a frequency of 400.0 Hz and the air is 23.0˚C?

2) A man hears a frequency of 320.0 Hz as a car pulls away from him at 22.0 m/s. If the air is 25.0˚C, what frequency does the driver hear?

3) A listener hears a train approach on a 20.0˚C day. He hears a frequency of 387 Hz while the engineer hears a frequency of 320.0 Hz. How fast is the train traveling?

4) A listener in front of an approaching train hears a frequency of 425 Hz while the engineer hears a frequency of 400.0 Hz. If the train is traveling 20.0 m/s, what must be the air temperature on this day?

Interactive Problem: 17.17

Resonance frequency of 425 Hz while the engineer hears a frequency of 400.0 Hz. If the train is traveling 20.0 m/s, what must be the air temperature on this day?

One vibrating object acts upon another to cause the second to vibrate as well.

Resonance is also referred to as sympathetic vibrations.

Resonance only occurs when the two objects have the same natural vibration rates-- there is no resonance if the objects have different natural frequencies, or if the source is not powerful enough!

A common and simple method of producing resonance is by using air columns:

Resonance in a closed tube occurs when sound waves are reflected so that waves entering the tube are 180˚ out of phase with the reflected part:

approx. (1/4)

The first point this occurs at is when half a wavelength is “folded” in half inside the tube!

Because the sound at the end of the tube is 3 dimensional the wavelength can be found using the following relationship:

d = diameter of tube

= 4(l + .4d)

l = length of tube

Other resonance points can be found using longer tubes:

approx. (3/4)

Resonance can also be found using a tube that is open at both ends:

approx. (1/2)

= 2(l + 0.8d)

With longer tubes, further resonance points could be found at lengths of 2/2, 3/2, 4/2, 5/2, etc.

1) A tuning fork is used to produce resonance in a closed tube of diameter 4.00 cm on a day when the air temperature is 19.3˚ C. The resonant tube is measured to be 28.0 cm long. What must have been the frequency of the fork?

2) A student doing a lab finds that a tuning fork of 256 Hz produced resonance in a closed tube that was 31.90 cm long and with a diameter of 4.00 cm. What was the speed of sound in the lab?

3) A 4.00 cm diameter open tube that is 38.20 cm long produces resonance with a tuning fork of 416 Hz. What is the air temperature?

4) A chimney of diameter 18.0 cm with the damper open acts as an open resonant tube on a day when the air in the chimney is 38.0°C. How long must the chimney be if the resonant frequency produced is 33.1 hz?

5) A physics student blows across the top of a graduated cylinder that is 23.0 cm tall and 4.00 cm in diameter and produces a resonant frequency of 350.0 hz. What is the temperature of the air in the cylinder?

Standing Waves and Harmonics as an open resonant tube on a day when the air in the chimney is 38.0°C. How long must the chimney be if the resonant frequency produced is 33.1 hz?

A wave that remains in a fixed position, as when plucking a string:

Consists of Loops

And Nodes

The Laws of Strings as an open resonant tube on a day when the air in the chimney is 38.0°C. How long must the chimney be if the resonant frequency produced is 33.1 hz?

The frequency of a plucked string is inversely proportional to the length and diameter of the string.

It is also inversely proportional to the square root of the density of the string.

It is directly related to the square root of the tension of the string.

f = l’• d’ • F • D’

f’ l • d • F’• D

1) A certain string will produce a frequency of 320 Hz when it is 70.0 cm long and under a tension of 25.0 N. If the tension is increased to 30.0 N and the string is lengthened to 90.0 cm, what is f ’ ?

2) A tension of 253 N on a guitar string will produce a fundamental frequency of 452 Hz. To what tension should the string be changed to if the guitarist wants a frequency of 440 Hz from that string?

3) A certain string produces a frequency of 250 Hz. What would be the frequency if we doubled the length, doubled the diameter, and halved both the tension and the density?

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