Today’s Lecture Shock Waves Doppler Shifts Review

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Today’s Lecture
Shock Waves
Doppler Shifts
Review
Shock Waves – Sonic Booms.
Source moving with Vsource = Vsound
( Mach 1 - equals the sound barrier )
Source moving with Vsource
(Mach 1.4 - supersonic)
> Vsound
Pressure waves can not get away from the source when the source is moving
through a medium at a speed about the speed of waves in the medium, e.g.
speed of sound.
In other words, even objects not emitting any waves at rest becomes a
source of shock waves at a supersonic speed.
Shock Waves - Mach Cone
u=v
v
u
Shock Waves - Mach Cone
v
a
A triangle in the cone:
a side a = T⋅v – the
travel distance of the
wave within one period.
θ
b
a T ⋅v v
sin θ = =
= <1
b T ⋅u u
u
The hypotenuse b = T⋅u
– the travel distance of
the moving object.
θ − the Mach angle
u
v
- Mach number
The sharper the Mach angle, the higher the speed of the object
Doppler effect – change in the wave frequency and/or wave length
due to motion of the source or the observer.
Case #1 – moving observer.
r
v
L
The train is moving at a speed v, and the
cars have a length L.
When the observer is at rest there are
v
f =
L
cars passing by every
second
v in general is the speed of the wave, e.g. sound,
each train car represents a wave crest
Doppler effect – change in the wave frequency and/or wave length due
to motion of the source or the observer.
Case #1 – moving observer.
r
v
L
The train is moving at a speed v, and the cars
have a length L.
When the observer is moving in the same
direction as the train at a speed u, the relative
speed is v
– u and there are
v −u
f '=
L
cars passing by every
second
Doppler effect – change in the wave frequency and/or wave length due
to motion of the source or the observer.
Case #1 – moving observer.
r
v
L
The train is moving at a speed v, and the cars
have a length L.
When the observer is moving in the direction
opposite to the train at a speed u, the relative
speed is v
+ u and there are
v+u
f '=
L
cars passing by every
second
r
v
λ
With “+” and “-” corresponding to the opposite
and same directions of motion, respectively.
+-
The wave length does not change. (of course!)
The general equation is
f '=
v±u
λ
The original frequency is
r
u
Therefore,
f =
v
λ
v±u
f '= f
= f (1 ± u / v)
v
Doppler effect.
Case #2 – moving source.
When the source moves and observer is at rest the relative
velocity of the wave crests with respect to the observer is just the
wave speed – it does not change. But the wave length does!
Stationary Sound Source
Car horn
large λ
small λ
low f
high f
Source moving with
Vsource < Vsound ( Mach 0.7 )
Case #2 – moving source.
Imagine splashing water with time
intervals T, and making a circular
wave every T seconds.
In the time intervals between two
consecutive splashes a circle travels
a distance λ = T⋅v, where v is the wave speed.
So the distance between the consecutive circles is λ.
Now imagine moving your hand a distance Δx during the time T. Then the distance
between the consecutive circles in front of your hand will be λ’ = λ – Δx.
And behind your hand the distance will be λ’ = λ + Δx.
Your hand is the source of waves here and its speed is u = Δx/T.
Therefore the distance between the circles is:
Doppler Effect – Moving Source
The source of the sound is moving with velocity u.
During one period, T, the
source moves uT.
The distance between wave
crests ahead of the source is:
A similar analysis shows that the distance between wave crests behind
the source is increased by the same fractional amount.
Case #2 – moving source.
u
λ ' = λ (1 ± )
v
The wavelength changes if the source of the
waves moves.
It decreases if the source and the wave move in
the same direction (approaching source)
The wavelength increases if the source moves in the direction
opposite to the wave motion (receding source), “+” in the equation.
What about the wave speed? does it change?
What about the frequency registered by an observer at rest?
u
λ ' = λ (1 ± )
v
What about the wave speed? does it change?
What about the frequency registered by an
observer at rest?
The wave speed does NOT change, since the circles, once
generated, loose any connection with the source, and cannot
“know” about motion of the source. They only care about the
mechanical properties of the medium.
Therefore the wave length and the frequency are connected by the
usual equations
Here again “-” is for approaching source – higher frequency.
“+” is for receding source – lower frequency.
General Expression for Doppler Shift
For a moving observer:
For a moving source:
Both of these expressions are included in the general relationship:
Here uo is the velocity of the observer and us is the velocity of the source.
It is usually best to use physical intuition to determine the signs. The
Doppler shift results in an increase in the frequency when the object and
the source are approaching each other with an increase in velocity.
Police Radar – sends frequency f.
The car moving toward the radar (moving observer) at a
speed u receives a frequency:
v+u
f '=
f
v
It reflects the received frequency f’, but it is a source moving
toward the radar, so that the radar receives a frequency
f'
1+ u / v
= f
f ''=
1− u / v
1− u / v
If u/v is small, which is typical for electromagnetic waves (v =
light speed).
f ' ' = f (1 + 2u / v )
The frequency of the received signal is higher by a fraction
2u
v
Why the equations for moving source
u
f
λ ' = λ (1 ± ) f ' =
1± u / v
v
and moving observer are different ?
f ' = f (1 ± u / v)
Isn’t all motion relative?
There is third player in the game – the medium for the waves.
It makes a lot of difference, whether or not the source of the
wave moves with respect to the material medium.
The mechanical waves (surface waves, string waves, sound etc.)
absolutely need a material medium to propagate.
Are there any waves that do not need a material medium?
Electromagnetic waves can propagate in vacuum and do not
require any medium. Does it make any difference?
A lot of difference! The whole theory of special relativity can be
derived from it.
For light and other electromagnetic waves only the relative velocity
of the source and the observer matter.
If this velocity is much lower than the speed of light
(the normal situation), the equations become:
The upper sign is for motion away from each other;
the light becomes more red.
The lower sign is for motion toward each other;
the light becomes more violet.
Note that the expression for the Doppler shift is the geometric mean and u<c.
Waves from a source, which is receding from us, are perceived to
have lower frequency and larger wavelength.
Red shift (shift toward larger wavelengths) in the spectra of
distant galaxies is an evidence of expanding universe.
Sound Waves in Review
Sound waves are longitudinal waves
propagating in gases, liquids, or solids.
The velocity of sound is given by
The intensity of sound depends on
the square of the pressure fluctuations
or the displacement velocity of the air.
Audible sound covers many orders of
magnitude. By convention the measure
of intensity is called decibel level:
Doppler effect for moving source
and moving observer:
Upper sign approaching
Lower sign receding.
Summary of Waves
A periodic continuous wave is
characterized by its wavelength
and period.
A simple harmonic wave can
be written as
Newton’s laws show that for small
amplitude waves on a string
Waves carry energy characterized
by intensity or flux that falls off as 1/r2
Many waves of small amplitude satisfy
the linear wave equation and those that
do obey the superposition principle.
Speed of Sound in a Gas
γP
v=
ρ
Assume that you have Argon gas an au of 40. It is to be
used in a moderate vacuum with a pressure of P=100Pa
at room temperature, 20oC. What is the speed of sound
in this gas?
One mole of this gas weighs 40g. The volume for one mole can be found
from the ideal gas law:
For an inert gas γ=5/3 and
If you know the temperature then we found during the previous lecture.
Wave Intensity
A wire with a linear density m = 5.0g/m is under 450N of tension. What is the
maximum average power carried by transverse waves so that the wave
amplitude does not exceed 10% of the wavelength?
Since ω = vk = 2πv/λ, this expression becomes:
From the given conditions A=.1λ and:
Decibel Levels
Number of dB, decibel, dimensionless parameter
I 0 = 10
−12
W/m
2
⎛I ⎞
β = 10 log ⎜⎜ ⎟⎟
⎝ I0 ⎠
I0 is the threshold of sensitivity of a normal ear.
However decibels are typically concerned with
relative changes in sound intensity.
Another example: (a) A rock band turns up their amplifiers so that the
decibel level is increased from 80dB to 96dB. What is the relative
change in sound intensity?
(b) What is the actual sound intensity when β=96db?
In general: sound level is going up/down by x dB sound intensity is multiplied by 10 ± x / 10
Doppler Shift
A wave source approaches you while you measure a frequency f1. As the
source recedes you measure a frequency f2. (a) If the speed of the wave is v,
find the expression for the source speed u and (b) the stationary frequency f.
As the source is
approaching:
Solving for u/v:
Solving for the
stationary frequency f:
As the source
is receding:
Carnot Cycle
What is the net work and
efficiency of a Carnot cycle?
Wtotal = WAB + WCD
A
B
D
C
Since ThVBγ−1=TcVCγ-1
and ThVAγ-1=TcVDγ-1
Qc Tc
=
Qh Th
Efficiency
We have VB / VA = VC / VD and
W Qh − Qc Th − Tc
=
=
e=
Qh
Qh
Th
What is the net work of
a Carnot cycle?
Carnot Cycle
Wtotal = WAB + WCD
A
B
D
Since ThVBγ−1=TcVCγ-1
and ThVAγ-1=TcVDγ-1
C
We have VB / VA = VC / VD and
All we need to find the net work is the expansion ratio and ΔT.
Carnot Refrigerator
Isothermal
contraction
adiabatic
expansion
adiabatic contraction
Isothermal
expansion
The key part is the isothermal process DC – the only time the system
is in contact with the cold reservoir. The gas is made to expand taking
heat Qc from the cold reservoir.
Carnot Refrigator
A home freezer, working between
-18°C inside and 30°C outside has
Inverting the cycle building a refrigerator
How much electrical energy is required to cool 0.53kg of water by 1°C?
How much heat will be rejected to the air in the kitchen?
Another Thermo Cycle
A
B
D
C
(a) Find the expressions and quantity for the net
work done by and net heat transferred to a
monotonic gas in the cycle ABCD as shown.
First define ΔV = VB – VA = VC –VD
and ΔP = PA – PD = PB – PC.
The net work done is the area contained
inside the rectangle which is:
The net heat transferred to the gas is found from:
Another Thermo Cycle
A
B
D
C
(b) Find the efficiency of the cycle when the
working fluid is a monotonic gas.
The heat added to the cycle is:
The efficiency is:
(c) Compare this efficiency with that of a Carnot Cycle.
How does the efficiency change for this cycle with a diatomic gas?
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