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Physics 1251 The Science and Technology of Musical Sound. Unit 2 Session 16 MWF The Wave Properties: Propagation. Physics 1251 Unit 2 Session 16 Wave Properties: Propagation. Foolscap Quiz:

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### Physics 1251The Science and Technology of Musical Sound

Unit 2

Session 16 MWF

The Wave Properties: Propagation

Foolscap Quiz:

What is frequency of the pitch that a person will perceive, if he hears the harmonics in the vibration recipe: 350 Hz, 525 Hz, 700 Hz and 875 Hz?

Answer:175 Hz (F3), fundamental of series:

350 Hz (F4), 525 Hz (C5), 700 Hz (F5) and

875 Hz (A5).

We hear the “difference” frequencies:

(525-350) =(700-525) =(875-350) = 175 Hz!

Georg Phillip Telemann

Sonata IV in E minor

Recorded by Ron DiIulio (UNT September 2001)

Georg Phillip Telemann

Sonata IV in E minor

Notice the “difference tones”

1′ Lecture:

• The intensity of sound decreases with the square of the inverse of the distance from the source in an open area.

I / I0 = (r0 /r) 2

1′ Lecture:

• The Sound Intensity Level (SIL) diminishes with distance thus:

SIL= SIL0 – 20 Log (r/ r0 )

Thought question:

What happens to a sound wave as it propagates or travels?

• Wave is absorbed.

• not much: ~ 10 dB/km

• depends of geometry.

The intensity of a spherical sound wave decreases as the distance from the source increases. [“Inverse”]

d [m] I [W/m2]

1.0 1.0

=12 /12

=12/2 2

=12 /3 2

= 12 /4 2

= 12 /5 2

As 1/r 2

2.0 0.25

3.0 0.11

4.0 0.068

5.0 0.040

Intensity is Power per Unit Area

Why does intensity

diminish as 1/r 2?Area = 2/3π ‧ r 2

I = Power/Area

I = I0 (A0 /A)

I= I0 (r0 / r ) 2

A = ⅔π r 2

r

Demonstration of Inverse Square Law:

I1 I1‧ (r1 /r2) 2 I1 ‧ (r1 /r3)2

Source

r1

r2

r3

The fraction of the sound energy that enters the ear decreases as the ratio of the inverse of the square of the distance because the fractional area decreases as 1/r 2.

80/20Inverse Square Law:

The intensity of sound (originating from a point source in an open environment) diminishes as the square of the inverse ratio of the distances from a source.

I/ I0= (r0 /r) 2

Practice:

The sound intensity is 1.0 x10 -9 W/m2 at a distance of 3.0 m from a speaker. What is the intensity 20 m away?

I2 / I1 = ( r1 /r2 ) 2 ;

I2 / (1 .0 x10-9 W/m2) =( 3.0 m / 20 m) 2

I2 = (1 .0 x10-9 W/m2) (.0225) =2.25 x10 -11 W/m2

80/20Inverse Square Law:

The Sound Intensity Level (SIL) decreases by 20 dB for every 10x increase in distance.

SIL= SIL0 – 20 Log (r/ r0)

10 Log(I/I Propagationthreshold) – 10 Log(I0 /Ithreshold) = 20 Log((r0 /r))

SIL – SIL0 = -20 Log(r/r0 )

Physics 1251 Unit 2 Session 16 Wave Properties: Propagation

80/20Inverse Square Law:

If I/ I0= (r0 /r) 2

Then 10 Log (I/ I0 )= 10 Log((r0 /r) 2)

10 Log(I/Ithreshold) – 10 Log(I0 /Ithreshold) = 20 Log((r0 /r))

SIL – SIL0 = -20 Log(r/r0 )

SIL= SIL0 – 20 Log (r/ r0)

Practice:

The sound intensity level of the professor’s lecture is 80 dB on the front row of the classroom, 3 m from the his mouth. What is the SIL at the back row if it is 6 m away and there are no reflections?

SIL = SIL0 – 20 dB Log ( r/r0 )

SIL = 80 dB – 20 dB Log ( 6 m/3 m) = 74 dB

Summary

• The intensity of sound decreases with the square of the inverse of the distance from the source in an open area.

I / I0 = ( r /r0 ) 2

• The Sound Intensity Level diminishes with distance thus

SIL= SIL0 – 20 Log (r/ r0)