Sound wave – longitudinal wave of compressions and rarefactions.
What are those black circles?
divide lines
Pressure
Speeds:
of molecular motion;
of the collective motion of the air;
of the sound wave.
Normal conversation – speed of collective motion 45 µ m/s. Amplitude – 10 nm.
displacement
Speed of a wave on a string
v=
F
µ
•
Tension of the string, F, provides the restoring force
•
Mass per unit length, µ = m/L, measures inertia of the string, that slows down any wave propagation Speed of sound ?
v=
B
ρ
•
Bulk modulus of elasticity, B, defines the restoring force
•
Density of the medium, ρ , measures inertia of the medium that slows down any wave propagation ∆P
B=−
∆V / V
∆ V/V is the fractional change of volume (can be measured in %)
Large bulk modulus of elasticity corresponds to a large change of pressure – vigorous restoring force – at small fractional change of volume
v=
B
ρ
∆P
B=−
∆V / V
• A hard material with a low compressibility and low density ­ Aluminum ­ high speed of sound.
• A soft material with a high density – Lead – low speed of sound.
Speed of sound in a gas
v=
B
ρ
Can we calculate the bulk modulus of elasticity, B, of a gas?
What process should we assume?
∆P
B=−
∆V / V
Sound waves are propagating quickly.
No time for heat exchange!
⇒ Adiabatic process.
γ
PV = c = const
P = cV
dP
P
−γ −1
= −γ cV
= −γ
dV
V
−γ
Speed of sound in a gas
v=
B
ρ
∆P
dP
B=−
= −V
∆V / V
dV
dP
P
= −γ
dV
V
γP
v=
ρ
B = γP
γ = Cp/Cv – the constant for an adiabatic process in the gas
P pressure of the gas
ρ ­ density of the gas
Speeds of sound for different gases at normal conditions
γP
v=
ρ
γ = Cp/Cv – the constant for an adiabatic process in the gas
P pressure of the gas
ρ ­ density of the gas
For two different gases, the ratio of the sound speeds can be calculated as
γ 2P
γ 1P
v2 / v1 =
/
ρ2
ρ1
If the two gases are at the same pressure, P:
γ2
ρ1
v2 / v1 =
⋅
γ1
ρ2
Example: air and Helium at normal conditions
Ratio of the sound speeds:
γ2
ρ1
γ2
M1
v2 / v1 =
⋅
=
⋅
γ1
ρ2
γ1
M2
The volume occupied by one mole of a gas at the normal conditions is the same for all gases – 22.4 liters. Therefore, the ratio of densities is equal to the ratio of molar weights. (1) Air – a mixture of N2 ( M = 28 g/mole) and O2 ( M = 32 g/mole) Average molar weight – M1 = Mair = 29 g/mole.
Composed of diatomic molecules, γ
1
= γ
air
= 1.4.
(2) Helium M2 = MHe = 4 g/mole. A monatomic molecule, γ 2 = γ
1.67.
1.67 29
In air, v1 = 343 m/s.
v2 / v1 =
1.4
⋅
4
= 2.94
He
= In Helium, v2 = 1008 m/s
Intensity of sound
I – average intensity
1
I = s0ω∆P0 s0 – the amplitude of displacement of the air
2
∆ P the amplitude of pressure variations
∆P
I=
2 ρv
2
0
Expressions making a bit more sense:
Quadratic in the amplitude of pressure variations – proportional to potential energy of the gas deformations.
1
2 2
2
I = ρ ωs0 v = ρ v osc v = 2 K ⋅ v
2
Quadratic in the amplitude of velocity of the oscillations of the gas and the gas density. Proportional to kinetic energy of the gas motion and the wave speed.
−6
I = 10 W/m , v = 343 m/s ⇒ v osc = 45 μm/s
2
Sensitivity of human ear..
Most sensitive at about 4000 Hz.
343 m/s
λ = v/ f =
=
4000 Hz
= 8.6 cm
Eardrum ­ about 1 cm in diameter