Lecture 13
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Transcript of Lecture 13
Waves and Oscilla-ons
Kepler planets
Lecture 13 – Sound Waves Textbook reference: 17.1-‐17.3
Last -me: Standing Waves
Standing waves have the form:
• They can be split into a spa-al and a temporal part.
• The par-cles that make up the medium undergo SHM with amplitude 2Asin(kx).
y = 2A sin(kx) cos(ωt)
Last -me: Standing Waves
Nodes: points on a standing wave with zero amplitude.
An*nodes: points on a standing wave at which maximum displacement occurs
fn =v
λn= n
v
2L=
n
2L
�T
µ“Natural frequencies” Fundamental: n=1
Demo via YouTube
Rubens Tube demo with more fire and good explana-on
Wa8: Standing Waves – vibra-on generator and wire
The different modes can be generated easily, and the dependence of the frequency on tension shown.
Sound waves
• Longitudinal waves • Can move in three dimensions
• Need a medium/material to travel through
• As sound waves travel through air they move the molecules that make up the air, crea-ng high and low pressure regions.
• Can be modeled as a sinusoidal wave (like on a string)
How to make a sound wave
Compression High Pressure
Rarefac-on Low Pressure
Assuming the piston is being moved with SHM
…and you thought we were done with pistons!
Sound Waves: Compression waves
• Compressible gas, ini-ally uniform density
• Piston suddenly moved to the right – Gas in front is compressed
• Piston comes to rest, but compression region con-nues to move – Corresponds to a longitudinal pulse
travelling through the tube with speed v
– Speed of the piston is not the same as the speed of the wave
Quan-ta-ve
Let be the posi-on of a small element rela-ve to its equilibrium posi-on.
Each element moves with SHM parallel to direc-on of the wave.
Displacement from equilibrium:
smax is the displacement amplitude.
We can use this to derive an expression for the pressure, which is also periodic.
s(x, t)
s(x, t) = smax cos(kx− ωt)
Pressure
Vi = A∆x
∆V = A∆s = A(s1 − s2)
A defini-on…Bulk Modulus
B ≡ volume stress
volume strain= −∆F/A
∆V/Vi= − ∆P
∆V/Vi
⇒ ∆P = −B∆V
Vi
Consider a parcel of gas with thickness Δx
Stress: force on a material Strain: what happens to the material under a given stress.
Pressure
⇒ ∆P = −BA∆s
A∆x
∆P = −B∂s
∂x
∆P = −B∂
∂x[smax cos(kx− ωt)]
∆P = Bsmaxk sin(kx− ωt)Using equa-on for s(x,t)
This is a Pressure Amplitude
Periodic Sound Waves: Pressure vs. Displacement
• A sound wave may be considered either a displacement wave or a pressure wave
• Since s (x, t) = smax cos (kx – ωt) and ΔP = Bsmax k sin (kx – ωt)
the pressure wave is 90o out of phase with the displacement wave – The pressure is a maximum when the
displacement is zero, etc.
If you blow across the top of an empty sof-‐drink bogle, a pulse of sound travels down through the air in the bogle. At the moment the pulse reaches the bogom of the bogle, what is the correct descrip-on of the displacement of elements of air from their equilibrium posi-ons and the pressure of the air at
this point?
1. The displacement and pressure are both at a maximum.
2. The displacement and pressure are both at a minimum.
3. The displacement is zero, and the pressure is a maximum.
4. The displacement is zero, and the pressure is a minimum.
If you blow across the top of an empty sof-‐drink bogle, a pulse of sound travels down through the air in the bogle. At the moment the pulse
reaches the bogom of the bogle, what is the correct descrip-on of the displacement of elements of air from their equilibrium posi-ons and the
pressure of the air at this point?
1. The displacement and pressure are both at a maximum.
2. The displacement and pressure are both at a minimum.
3. The displacement is zero, and the pressure is a maximum.
4. The displacement is zero, and the pressure is a minimum.
Because the bottom of the bottle is a rigid barrier, the displacement of elements of air at the bottom is zero. Thus, the pressure variation is at a minimum or a maximum. Since the pulse is moving downward, the pressure variation at the bottom is a maximum – see the diagram.
Let’s take a short break
Pressure from sound waves
The speed of sound
Ini-al State
Final State
Force applied on piston from outside is in equilibrium with force applied by gas on the piston.
The speed of sound
Afer -me interval every bit of gas in the element is moving to the right with speed . The speed of sound in this medium is .
∆t
vx
v
Impulse = ∆momentum
I =�
F∆t = (A∆P∆t)̂iSum of forces, using defini-on of pressure P=F/A
The speed of sound
∆P = −B∆V
Vi= −B
(−vxA∆t)
vA∆t= B
vxv
⇒ I = (ABvxv∆t)̂i
From 8 slides ago where we defined B
Volumes come from diagram here:
I =�
F∆t = (A∆P∆t)̂iImpulse from previous page
The speed of sound
⇒ I = (ABvxv∆t)̂i
∆p = m∆v = (ρVi)(vx̂i− 0) = (ρvvxA∆t)̂i
⇒ ABvxv∆t = ρvvxA∆t ⇒ v =
�B
ρ
Momentum of gas parcel. Impulse I causes a change in momentum Δp
The speed of sound (liquid or gas)
v =
�B
ρ
v =
�elastic property
inertial property
v =
�T
µon a string
Speed of Sound in Air
• The speed of sound also depends on the temperature of the medium – Par-cularly important with gases
• For air, the rela-onship between the speed and temperature is
– 331 m/s is the speed at 0o C – TC is the air temperature in Celsius
• Exercise for student: show that where TK is temperature in Kelvin
C(331 m/s) 1273 CTv = +
€
v = (331 m/s) TK273
The speed of sound
Pressure and displacement
∆Pmax = Bsmaxk = (ρv2)smax(ω
v) = ρvωsmax
More useful to relate the maximum change in pressure to density than to bulk modulus.
Pressure amplitude
Intensity of Periodic Sound Waves
Waves carry energy, sound waves also carry energy.
As a piston moves back and forward crea-ng sound waves, it is doing work on a gas.
The rate of work done gives us the power
W = F · x
Power = F · vx
Intensity of Periodic Sound Waves Power = F · vx
Power = [∆P (x, t)A]̂i · ∂
∂t[s(x, t)]̂i
Power = [ρvωAsmax sin(kx− ωt)]∂
∂t[smax cos(kx− ωt)]
Power = ρvωAsmax sin(kx− ωt)[ωsmax sin(kx− ωt)]Power = ρvω2As2max sin
2(kx− ωt)
Now we need to find the average power over one period. This is independent of x so we can choose any value of x, lets take x = 0. We start by finding the average value of the sin2(kx – ωt) term.
1
T
� T
0sin2(0− ωt)dt =
1
T
� T
0sin2(ωt)dt =
1
T(t
2+
sin 2ωt
2ω)T0 =
1
2
Intensity of Periodic Sound Waves
The average power is then given by:
Intensity is the power per unit area:
(Power)avg =1
2ρvω2As2max
I ≡ (Power)avgA
Intensity of Periodic Sound Waves
So for this case (the wave was moving in the x direc-on)
Or in terms of pressure (remember )
I =1
2ρv(ωsmax)
2
∆Pmax = ρvωsmax
I =(∆Pmax)2
2ρv
But sound waves can travel in 3 dimensions….
I =(Power)avg
4πr2
Homework Set 6: PHYS 1121: 6, 7, 8, 12, 13
PHYS 1131: 7, 9, 10, 12, 15, 16
This is an inverse-‐square law (happens all the -me in astrophysics!)
A vibra-ng guitar string makes very ligle sound if it is not mounted on the guitar body. Why does the sound have greater intensity if the string is agached to the
guitar body?
1. The string vibrates with more energy.
2. The energy leaves the guitar at a greater rate.
3. The sound power is spread over a larger area at the listener’s posi-on. 4. The sound power is concentrated over a smaller area at the listener’s
posi-on.
5. The speed of sound is higher in the material of the guitar body.
6. None of these answers is correct.
A vibra-ng guitar string makes very ligle sound if it is not mounted on the guitar body. Why does the sound have greater
intensity if the string is agached to the guitar body?
1. The string vibrates with more energy.
2. The energy leaves the guitar at a greater rate. 3. The sound power is spread over a larger area at the listener’s posi-on.
4. The sound power is concentrated over a smaller area at the listener’s posi-on.
5. The speed of sound is higher in the material of the guitar body.
6. None of these answers is correct.
The large area of the guitar body sets many elements of air into oscillation and allows the energy to leave the system by mechanical waves at a much larger rate than from the thin vibrating string.
Ques-on
A sound wave propagates in air at 27o C with frequency 4.00 kHz. It passes through a region where the temperature gradually changes and then moves through air at 0o C. Give numerical answers to the following ques-ons to the extent possible and state your reasoning about what happens to the wave physically.
(a) What happens to the speed of the wave? (b) What happens to the frequency? (c) What happens to the wavelength?
QUESTION!!!!!
The faintest sounds the human ear can detect at a frequency of 1000 Hz correspond to an intensity of about 1.00 x 10-‐12 W/m2, which is called the threshold of hearing. The loudest sound that the ear can tolerate at this frequency corresponds to an intensity of about 1.00 W/m2, the threshold of pain. Determine the pressure amplitude and displacement amplitude associated with these two limits.
ρair = 1.2 kg/m3 Sound speed cs = 343 m/s
Ques-on
A point source emits sound waves with an average power output of 80.0 W.
(a) Find the intensity 3.00 m from the source.
(b) Find the distance at which the intensity of the sound is 1.00 x 10-‐8 W/m2.