Standing waves
The interference of two sinusoidal waves of the same frequency and
amplitude, travel in opposite direction, produce a standing wave.
y1(x, t) = ymsin(kx - ωt),
y2(x, t) = ymsin(kx + ωt)
resultant wave: y'(x, t) = [2ymsin kx] cos(ωt)
nodes
Anti-nodes
y'(x, t) = [2ymsin kx] cos(wt)
If kx = nπ (n = 0, 1, …), we have y' = 0; these positions are
called nodes. x = nπ/k = nπ/(2π/λ) = n(λ/2)
y
kx
0
π/2
π
3π/2
2π
y'(x, t) = [2ymsin kx] cos(ωt)
If kx = (n + ½)π (n = 0, 1, …), y'm = 2ym (maximum);
these positions are called antinodes, x = (n + ½ )(λ/2)
y
kx
0
π/2
π
3π/2
2π
Standing waves and resonance
For a string clamped at both ends, at certain frequencies, the
interference between the forward wave and the reflected
wave produces a standing wave pattern. The string is said to
resonate at these certain frequencies, called resonance
frequencies.
L = l/2, f = v/λ = v/2L 1st harmonic
or fundamental mode
L = 2(λ/2 ), f = 2(v/2L) 2nd harmonic
f = n(v/2L), n = 1, 2, 3… nth harmonic
Reflection at a Boundary
Case a) String tied to wall.
– The reflected and incident pulses
must have opposite signs. A node is
generated at the end of the string.
Case b) String end can move.
– The reflected and incident pulses
must have same sign. An antinode
is generated at the end of the
string.
a
b
Reflection at a Boundary
Case a) String tied to wall.
– The reflected and incident pulses
must have opposite signs. A node is
generated at the end of the string.
Case b) String end can move.
– The reflected and incident pulses
must have same sign. An antinode
is generated at the end of the
string.
a
b
Reflection at a Boundary
Case a) String tied to wall.
– The reflected and incident pulses
must have opposite signs. A node is
generated at the end of the string.
Case b) String end can move.
– The reflected and incident pulses
must have same sign. An antinode
is generated at the end of the
string.
a
b
Sample Problem
A pulse is traveling down a wire and
encounters a heavier wire. What happens
to the reflected pulse?
1.
2.
3.
4.
The reflected wave changes phase
The reflected wave doesn’t change phase
There is no reflected wave
none of the above
Problem 16-84
A standing wave results from the sum of two transverse traveling
waves given by: y1(x, t) = 0.050 cos(πx - 4πt)
y2(x, t) = 0.050 cos(πx + 4πt)
where x, y1, and y2 are in meters and t is in seconds.
Note y1 and y2 are cosines, not sines – but y’ is *still* = y1 + y2
(trig identity)
where α = πx ‒ 4πt and β = πx + 4πt.
Alternatively, cos(α) = sin(α + π/2) = sin(α)cos(π/2) + sin(π/2)cos(α)
Problem 16-84
A standing wave results from the sum of two transverse traveling
waves given by: y1(x, t) = 0.050 cos(πx ‒ 4πt)
y2(x, t) = 0.050 cos(πx + 4πt)
where x, y1, and y2 are in meters and t is in seconds.
where α = πx ‒ 4πt and β = πx + 4πt.
Problem 16-84
A standing wave results from the sum of two transverse traveling
waves given by: y1(x, t) = 0.050 cos(πx ‒ 4πt)
y2(x, t) = 0.050 cos(πx + 4πt)
where x, y1, and y2 are in meters and t is in seconds.
Amplitude in space
Amplitude in time
Problem 16-84
A standing wave results from the sum of two transverse traveling
waves given by: y1(x, t) = 0.050 cos(πx ‒ 4πt)
y2(x, t) = 0.050 cos(πx + 4πt)
where x, y1, and y2 are in meters and t is in seconds.
A) Find smallest non-negative x where y’=0.
Amplitude of composite wave:
2ymcos(πx) = 0 => πx = π/2, 3π/2, 5π/2, ...
Smallest when x = 1/2
Problem 16-84
A standing wave results from the sum of two transverse traveling
waves given by: y1(x, t) = 0.050 cos(πx - 4πt)
y2(x, t) = 0.050 cos(πx + 4πt)
where x, y1, and y2 are in meters and t is in seconds.
B) Find times the particle at x=0 has zero velocity.
Take the temporal derivative of y’:
x=0
=1
sin(4πt) = 0 => 4πt = 0, π, 2π, 3π, ... => t = 0, 1/4, 1/2, 3/4, ...
Pipes
Open on both ends
Antinodes both ends.
{
Open on one end,
closed on the other end
node on closed end,
antinode on open end.
{
Chapter 17
Waves (Part II)
Sound wave is a longitudinal wave.
The speed of sound
Speed of a transverse wave on a stretched string
The speed of sound
Speed of a sound wave (longitudinal):
Speed of sound in gases & liquids is described by the
bulk modulus B ( = - Δp/(ΔV/V)) and density ρ:
Speed of sound:
air(0oC): 331 m/s, air (20oC): 343 m/s, water (20oC): 1482 m/s
The speed of sound
Speed of a sound wave (longitudinal):
Speed of sound in a solid is described by Young’s
modulus E and density ρ:
Speed of sound in aluminum: E = 7x1010 Pa,
ρ = 2.7x103 kg/m3 => v = 5100 m/s,
In general,
vsolid > vliquid > vgas
Traveling Sound Wave
To describe the sound wave, we use the displacement of an
element at position x and time t:
s(x, t) = smcos( kx – ωt )
sm: displacement amplitude
k = 2π/λ
ω = 2π/T = 2πf
As the wave moves, the air pressure at each point changes:
Δp(x, t) = Δpmsin( kx – ωt )
Pressure amplitude: Δpm = (vρω)sm
Interference
Two sound waves traveling in
the same direction, how they
would interfere at a point P
depends on the phase difference
between them.
ΔL = L1 – L2
If the two waves were in phase when they were emitted,
then the phase difference depends on path length
difference: ΔL = L1 – L2:
ϕ/2π = ΔL/λ
Interference
y
For distant points,
d
ΔL = d sinθ
θ
D
ΔL
Δϕ/2π = ΔL/λ
ΔL = d sinθ = mλ
y/D = tanθ
If Δϕ = m (2π)
or
ΔL/λ = m
( m = 0, 1, 2 …)
We have fully constructive interference ΔL = d sinθ = mλ
If Δϕ = (2m +1)π or
ΔL/λ = m + ½
( m = 0, 1, 2 …)
We have fully destructive interference d sinθ = (m + ½) λ
Problem 17-19
Two loudspeakers are located
3.35m apart. A listener is 18.3m
from one and 19.5m from the
other speaker.
y
d
θ
D
ΔL
A) What is the lowest audible frequency (20 Hz – 20 kHz) that gives a
minimum (fully destructive interference) signal at the listener’s location?
L1=18.3m, L2=19.5m
=> ΔL = 1.2m
We want the path difference to be an odd multiple of
the half wavelength. Or in other words,
ΔL = (m+1/2)λ,
where m=0, 1, 2, ...
Problem 17-19
Two loudspeakers are located
3.35m apart. A listener is 18.3m
from one and 19.5m from the
other speaker.
y
d
θ
D
ΔL
A) What is the lowest audible frequency (20Hz - 20kHz) that
gives a minimum signal at the listener’s location?
L1=18.3m, L2=19.5m
=> ΔL = 1.2m
We want ΔL = (m+1/2)λ,
vsound = f λ
where m=0, 1, 2, ...
=> flowest happens when λ is the largest
Problem 17-19
Two loudspeakers are located
3.35m apart. A listener is 18.3m
from one and 19.5m from the
other speaker.
y
d
q
D
DL
A) What is the lowest audible frequency (20Hz - 20kHz) that
gives a minimum signal at the listener’s location?
L1=18.3m, L2=19.5m
We want DL = (n+1/2)l,
=> DL = 1.2m
where n=0, 1, 2, ...
Problem 17-19
Two loudspeakers are located
3.35m apart. A listener is 18.3m
from one and 19.5m from the
other speaker.
y
d
q
D
DL
B) What is the second lowest audible frequency (20Hz - 20kHz)
that gives a minimum signal at the listener’s location?
L1=18.3m, L2=19.5m
=> ΔL = 1.2m
We want ΔL = (n+1/2)λ,
where n=0, 1, 2, ...
Problem 17-19
Two loudspeakers are located
3.35m apart. A listener is 18.3m
from one and 19.5m from the
other speaker.
y
d
q
D
DL
D) What is the lowest audible frequency (20Hz - 20kHz) that
gives a maximum signal at the listener’s location?
L1=18.3m, L2=19.5m
We want ΔL = nλ,
=> ΔL = 1.2m
where n=1, 2, ...
Sample Problem
Today, a lightning bolt is seen in the distance. You start
counting when you saw the flash. Twelve seconds elapsed
until you heard the thunder. How far away was the lightning?
1) 4.1km
2) 1200m
4) Yikes, head for cover
3) 343m
5) none of the above
Daily Quiz-Lightning, March 27, 2006
Today, a lightning bolt is seen in the
distance. You start counting when you
saw the flash. Twelve seconds elapsed
until you heard the thunder. How far
away was the lightning?
d = vsoundt = (343m/s)(12s)
= 4.1km = 2.5miles
1) 4.1km
2) 1200m
4) Yikes, head for cover
3) 343m
5) none of the above
Intensity and Sound Level
Intensity of a sound wave at a surface is defined as the
average power per unit area
I = P/A (Watts/m2)
For a wave s(x, t) = smcos( kx – ωt )
I = ½ ρvω2s2m
v = ω/k
Variation of intensity with distance.
A point source with power Ps
I = Ps/4πr2
Ps – a property of the
source (W, or J/s)
Spherical surface area:
4πr2
The figure indicates three small patches 1,
2, and 3 that lie on the surfaces of two
imaginary spheres; the spheres are
centered on an isotropic point source S of
sound.
A) Rank the patches according to the intensity
of the sound on them, greatest first.
I = Ps
r1 = r2 => I1 = I2
/4πr2
r3 > r1 => I3 < I1
B) If the rates at which energy is transmitted through the three
patches by the sound waves are equal, rank the patches according to
their area, greatest first.
r = r => A = A
1
I = Ps/A
2
1
2
r3 > r1 => A3 > A1
The decibel Scale
The deci(bel) scale is named after Alexander Graham Bell
Sound level b is defined as
β = (10 dB)log (I/I0)
Unit for β: dB(decibel)
I0 is a standard reference intensity, I0 = 10-12 W/m2
Threshold of hearing: 10-12W/m2
Threshold of pain (depends on the type of music): 1.0W/m2
An increase of 10 dB => sound intensity multiplied by 10.
• The decibel scale
– Sound intensity range for human ear: 10-12 – 1 W/m2
– More convenient to use logarithm for such a big range.
– Sound level b is defined as
b = (10 dB)log10(I/I0)
• Unit for b: dB(decibel)
I0 is a standard reference intensity, I0 = 10-12 W/m2
Examples:
at hearing threshold: I = I0, b = 10 log101 = 0,
normal conversation: I ~ 106 I0, b = 10 log10106 = 60 dB
Rock concert: I = 0.1 W/m2,
b = 10 log10(0.1/10-12) = 110dB
Beats
Two sound waves with
frequencies f1 and f2 ( f1 ~ f2 )
reach a detector, the intensity
of the combined sound wave
vary at beat frequency fb
fb = f1 – f2
The Doppler effect
When the source or the detector is moving relative to the
media, a frequency which is different from the emitted
frequency is detected.
This is called Doppler effect.
• Case 1: source stationary, the
detector is moving towards
the source:
The detected frequency is:
Note: f ' is greater than f.
Case 2: source stationary, detector moving away
from the source:
f ' is smaller than f.
Cases 3 & 4: source moving, detector stationary
Source is moving toward detector, “-” sign,
f' >f
Source is moving away from detector, “+” sign,
f'<f
General formula for Doppler effect
v: speed of sound
vD: detector speed
vS: source speed
Rules:
Moving towards each other: frequency increases
Moving away from each other: frequency decreases
For numerator,
if detector is moving towards the source,
“+”
if detector is moving away from the source, “ - ”
For denominator,
if source is moving away from the detector, “ + ”
if source is moving towards the detector,
“ -”