1. Chapter 19 Sound Waves Molecules obey SHM s(x,t)=smcos(kx-t)
LO1 Jan 21/02
Chapter Sound Waves
Molecules obey SHM s(x,t)=smcos(kx-t)
leads to variations in density and pressure in the air
pressure is large where density is large but displacement s(x,t) is small
hence pressure(or density) wave
is 900 out of phase with displacement
p=p0 + pm cos(kx-t-/2) = p0 + pm sin(kx-t) (p is change in pressure relative to equilibrium)
= 0 + m sin(kx-t) “rho”
m = pm (0/B)
B is bulk modulus

2. Speed of Sound B=-p/(V/V) Water: B~2.2x109N/m2 Air: B~ 105 N/m2
gases are more compressible (B is smaller)
speed of sound in a fluid depends on the density and the elasticity-> i.e. the medium
v = (B/0)1/2 wave speed in a fluid
vair ~ 343 m/s at room temperature
vwater ~ 1482 m/s ( B is larger!)
vsteel ~ 5941 m/s

3. Intensity Level and Loudness
I=Pav/area
area of sphere = 4r => I  1/r2
sensation of loudness is not directly proportional to intensity
varies as log(I) eg. log(100)=2log(10)
define sound level SL=10 log(I/I0) in decibels (dB)
I0 = W/m2 is the hearing threshold
hence for I=I0 , SL= 10 log(1) = 0
pain threshold is I= 1 W/m2 or SL=10log(1012)=120 dB

5. Problem
When a pin of mass 0.1 g is dropped from a height of 1 m, 0.05% of its energy is converted into a sound pulse with a duration of 0.1 s.
(a) Estimate the range at which the dropped pin can be heard if the minimum audible intensity is W/m2 .
(b) Your result in (a) is much too large in practice because of background noise. If you assume that the intensity must be at least 10-8 W/m2 for the sound to be heard, estimate the range at which the dropped pin can be heard (In both parts, assume that the intensity is P/4r2 .)
(a) Sound energy is 5 x 10-4 (mgh) = 5 x 10-4 (1m)(10-4 kg)(9.8 m/s2 ) = 4.9 x 10-7 J
Pav = E/t = 4.9 x10-6 W = 4r2 x10-11 W => r ~ 200 m.
(b) r ~ 200/ (1000)1/2 = 6.24 m.

6. Problem
The equation I = Pav / 4r2 is predicated on the assumption that the transmitting medium does not absorb any energy.
It is known that absorption of sound by dry air results in a decrease of intensity of approximately 8 dB/km.
The intensity of sound at a distance of 120 m from a jet engine is 130 dB.
Find the intensity at 2.4 km from the jet engine (a) assuming no absorption of sound by air, and (b) assuming a diminution of 8 dB/km. (Assume that the sound radiates uniformly in all directions.)
(a) SL = 10 log(I1/I0)-10 log (I2/I0)= 10 log(I1/I2) = 10log(r22/r12)=20log(r2/r1)
20 log (20) = 26; SL at 2.4 km = ( ) dB = 104 dB
(b) Subtract 2.28 x 8 dB from result of (a)
SL= ( ) dB = 85.8 dB

7. Interference of Sound Waves
ripple

8. Interference and Phase Difference
Consider shifting one of the waves by a distance  relative to the other
they add up constructively
corresponds to a phase shift by  = 2
constructive interference for  = m2
or path difference L= m 
a shift by  =  corresponds to a shift of one of the waves by  /2
destructive interference for  = (2m+1)
or path difference L= (2m+1)  /2

9. Musical Sounds oscillating strings membranes air columns
standing wave patterns correspond to resonances
large amplitude oscillations push surrounding air and generate sound waves at the same frequency

10. Pressure waves in open pipe
Pressure waves in pipe
closed at one end