Intensity
Intensity = (average power)/(unit area)
Lecture 9
I = <P>/A = <F"y>/A = <p(x,t)"y(x,t)>
Intensity
But
p(x,t)"y(x,t) = BkA sin(kx – #t) #A sin(kx – #t)
= B#kA2 sin2(kx – #t)
and
Interference
So
I=
or
I=
1
2
B"kA 2 =
2
max
1
2
2
max
#B" 2 A 2 Note: <x> means
"p
p
p2
=
= max
2B
2 #" 2 #B
average of x.
<sin2(kx – #t)> = 1/2
!
1
§16.3
2
!
Example 16.8
Intensity (2)
We want I = 1Wm–2 at 20m from speakers in all
directions. What speaker power is needed?
• Solution: “all directions” means hemisphere
above ground, with area 1/2 x 4!(20m)2.
• For a point sound source, I = P/4!r2
“Inverse
• Sound intensity level often measured in
decibels: B = 10dB log10 (I/Io)
where I0 = 10–12Wm–2 is a reference intensity.
So acoustic power = 1 Wm–2 x 2513m2 = 2.5kW;
I0 is roughly the softest audible sound @ 1000 Hz
actually need more electrical power than this since
speakers are inefficient.
• B = 0 dB when I = I0 , and B = 120 dB
when I = 1Wm–2.
• Doubling intensity ! 3dB increase
§16.3
§16.3
3
4
Interference in Travelling
Sound Waves
Example 16.10
By how many dB does the intensity drop
when you move twice as far away from a
singing bird?
Solution:
The difference is given by B2–B1
•
= 10dB [log(I2/I0)–log(I1/I0)]
= 10dB log(I2/I1) = 10dB log(r12/r22)
= 10dB log[r12/(2r1)2]
= 10dB log(1/4) = –6.0dB
•! If two sources of produce identical sound signals in-phase
(coherent) then depending on difference in path lengths
from sources to listener, constructive (loud sound) or
destructive (soft sound) interference may occur.
•! If path difference d is whole
number of wavelengths d = n$,
then constructive. If path
difference is half integer
d = (n +1/2)$ then destructive.
!
Note: Also works for 2 light sources
e.g. Young’s double slit
§16.3
5
square law” similar to light.
6
§16.6
Interference in Travelling
Sound Waves
•! Here we DON’T have standing waves. Both
waves travel in roughly the same direction
•! Interference of travelling waves from two
identical sources, but:
–!Energy flow is channelled
–!Pressure & displacement have the same nodes
–!Nodes where paths differ by d = (n +1/2)$
(or d = m$/2 for odd m) (destructive interference)
–!Antinodes where paths differ by d = n$
(constructive interference)
§16.6
7
Example 16.14
Speakers A & B emit in-phase
sinusoidal waves. v = 350 ms-1.
Find what frequencies yield
a) constructive & b) destructive interference at
point P
• Solution: 1st find difference between path
lengths AP & BP.
AP = "(22+42) ! 4.47m, BP = "(12+42) !
4.12m.
Path diff d = 4.47 - 4.12 = 0.35m §16.6
8
Example 16.14 (2)
• Constructive (loud sound) when d = n$ = nv/f;
or f = nv/d
f = n x 350/0.35 = 1000, 2000, 3000 Hz, ….
• Destructive (soft sound) when d = (n+1/2)$
= (n+1/2)v/f; or f = (n+1/2) v/d
f = (n+1/2) x 350/0.35 = 500, 1500, 2500 Hz,
….
9
10
Speed of sound in gases
Beats
(not examinable)
• “Beats” occur when two sounds have almost the
same frequency. “Beats” means periodically varying amplitude
• Mathematically:
• Recall in Lecture 5 we derived the speed of
sound in a fluid
Asin#1t – Asin#2t = 2A sin[(#1–#2)t/2] cos[(#1+#2)t/2]
Beat Envelope High frequency
v=
• Result is an
amplitude
that varies
at the beat
frequency
ƒbeat = ƒ1–ƒ2
where B is the bulk modulus (Pa) and % is the
mass density (kg/m3) of the fluid.
!
§16.7
11
B
"
12
Doppler effect
• Gases are more complicated, because the
bulk modulus of a gas depends on the
pressure.
• Examples: moving sources (trains, etc.)
• Define all " as +ve when moving from L towards S.
• Because the density also depends on the
pressure, it turns out that the speed of sound
in a gas depends on the temperature and the
molar mass
!
v∝
T
M
§16.8
13
14