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Week 11: Sound

 

465

 

 

 

 

 

 

Source of Sound

P0 (Pa)

dB

 

 

All sounds beyond this point

 

 

 

 

are nonlinear shock waves

 

 

 

 

 

 

 

 

 

Human Death from Shock Wave Alone

200,000

200

 

 

1 Ton of TNT

632,000

210

 

 

Largest Conventional Bombs

2,000,000

220

 

 

1000 Tons of TNT

6,320,000

230

 

 

20 Kiloton Nuclear Bomb

63,200,000

250

 

 

57 Megaton (Largest) Nuclear Bomb

2,000,000,000

280

 

 

Krakatoa Volcanic Explosion (1883 C.E.)

63,200,000,000

310

 

 

Tambora Volcanic Explosion (1815 C.E.)

200,000,000,000

320

 

 

 

 

 

 

Table 6: Table of (approximate) P0 and sound pressure levels in decibels relative to the threshold of human hearing at 1012 watts/m2 of shock waves, events that produce distorted overpressures greater than one atmosphere. These “sounds” can be quite extreme! The Krakatoa explosion cracked a 1 foot thick concrete wall 300 miles away, was heard 3100 miles away, ejected 4 cubic miles of the earth, and created an audible pressure antinode on the opposite side of the earth. Tambora ejected 36 cubic miles of the earth and was equivalent to a 14 gigaton nuclear explosion (14,000 1 megaton nuclear bombs)!

11.4: Doppler Shift

Everybody has heard the doppler shift in action. It is the rise (or fall) in frequency observed when a source/receiver pair approach (or recede) from one another. In this section we will derive expressions for the doppler shift for moving source and moving receiver.

11.4.1: Moving Source

 

 

λ0

Source

Receiver

vs

 

vsT

λ

Figure 134: Waves from a source moving towards a stationary receiver have a foreshortened wavelength because the source moves in to the wave it produces. The key to getting the frequency shift is to recognize that the new (shifted) wavelength is λ= λ0 − vsT where T is the unshifted period of the source.

Suppose your receiver (ear) is stationary, while a source of harmonic sound waves at fixed frequency f0 is approaching you. As the waves are emitted by the source they have a fixed wavelength λ0 = va/f0 = vaT and expand spherically from the point where the source was at the time the wavefront was emitted.

However, that point moves in the direction of the receiver. In the time between wavefronts (one period T ) the source moves a distance vsT . The shifted distance between successive wavefronts in the direction of motion (λ) can easily be determined from an examination of figure 134 above:

λ= λ0 − vsT

(979)

We would really like the frequency of the doppler shifted sound. We can easily find this by using

f = f0(1 vr ) va

466

λ= va/f and λ = va/f . We substitute and use f = 1/T :

va

=

va − vs

f

f0

 

then we factor to get:

f =

f0

1

vs

 

va

 

Week 11: Sound

(980)

(981)

If the source is moving away from the receiver, everything is the same except now the wavelength is shifted to be bigger and the frequency smaller (as one would expect from changing the sign on the velocity):

f =

f0

(982)

1 +

vs

 

 

va

 

 

 

 

11.4.2: Moving Receiver

λ 0

Source

Receiver

vr

vaT’ vrT’

Figure 135: Waves from a stationary source are picked up by a moving receiver. They have a shortened period because the receiver doesn’t wait for the next wavefront to reach it, at receives it when it has only moved part of a wavelength forward. The key to getting the frequency shift is to recognize that the sum of the distance travelled by the wave and the receiver in a new period T must equal the original unshifted wavelength.

Now imagine that the source of waves at frequency f0 is stationary but the receiver is moving towards the source. The source is thus surrounded by spherical wavefronts a distance λ0 = vaT apart. At t = 0 the receiver crosses one of them. At a time T later, it has moved a distance d = vr T in

the direction of the source, and the wave from the source has moved a distance D = vaT

toward

the receiver, and the receiver encounters the next wave front.

 

This can be visualized in figure 135 above. From it we can easily get:

 

λ0

=

d + D

(983)

 

=

vr T + vaT

(984)

 

=

(vr + va)T

(985)

vaT

=

(vr + va)T

(986)

We use f0 = 1/T , f = 1/T (where T

is the apparent time between wavefronts to the receiver)

and rearrange this into:

 

 

vr

 

 

f = f0

(1 +

)

(987)

 

 

 

 

va

 

Again, if the receiver is moving away from the source, everything is the same but the sign of vr , so one gets:

(988)