These notes cover the first block in vibration and wave physics.
You track how displacement links to pressure, density, speed, and energy.
The focus is longitudinal motion, first in gases, then in solids.
Part I, Sound Waves in Gases
1. Core Variables
For small disturbances, split each quantity into equilibrium plus perturbation.
P = P 0 + p , V = V 0 + v , ρ = ρ 0 + ρ d . P = P_0 + p, \quad V = V_0 + v, \quad \rho = \rho_0 + \rho_d. P = P 0 + p , V = V 0 + v , ρ = ρ 0 + ρ d .
Use η ( x , t ) \eta(x,t) η ( x , t ) x x x
δ = Δ V V 0 = ∂ η ∂ x . \delta = \frac{\Delta V}{V_0} = \frac{\partial \eta}{\partial x}. δ = V 0 Δ V = ∂ x ∂ η .
Define condensation, positive in compression, as
s = Δ ρ ρ 0 = ρ d ρ 0 . s = \frac{\Delta \rho}{\rho_0} = \frac{\rho_d}{\rho_0}. s = ρ 0 Δ ρ = ρ 0 ρ d .
Mass stays fixed for each small fluid element:
ρ V = ρ 0 V 0 . \rho V = \rho_0 V_0. ρ V = ρ 0 V 0 .
Write ρ = ρ 0 ( 1 + s ) \rho = \rho_0(1+s) ρ = ρ 0 ( 1 + s ) V = V 0 ( 1 + δ ) V = V_0(1+\delta) V = V 0 ( 1 + δ )
( 1 + s ) ( 1 + δ ) = 1. (1+s)(1+\delta) = 1. ( 1 + s ) ( 1 + δ ) = 1.
For small signals, drop the second order term s δ s\delta sδ
s = − δ = − ∂ η ∂ x . \boxed{s = -\delta = -\frac{\partial \eta}{\partial x}}. s = − δ = − ∂ x ∂ η .
2. Elastic Response and Sound Speed
Bulk modulus B B B
B ≡ − V ∂ P ∂ V . B \equiv -V\frac{\partial P}{\partial V}. B ≡ − V ∂ V ∂ P .
In linear acoustics:
p = − B δ = B s . p = -B\delta = Bs. p = − B δ = B s .
For ideal gases:
Isothermal disturbance, B = P 0 B = P_0 B = P 0
Adiabatic disturbance, B = γ P 0 B = \gamma P_0 B = γ P 0
Sound speed then follows:
c 2 = B ρ 0 , c = B ρ 0 . c^2 = \frac{B}{\rho_0}, \qquad c = \sqrt{\frac{B}{\rho_0}}. c 2 = ρ 0 B , c = ρ 0 B .
For adiabatic sound in an ideal gas:
c = γ P 0 ρ 0 . c = \sqrt{\frac{\gamma P_0}{\rho_0}}. c = ρ 0 γ P 0 .
Adiabatic changes satisfy
P V γ = constant . PV^\gamma = \text{constant}. P V γ = constant .
Differentiate this relation:
d P V γ + γ P V γ − 1 d V = 0 , dP\,V^\gamma + \gamma P V^{\gamma-1} dV = 0, d P V γ + γ P V γ − 1 d V = 0 ,
so
− V d P d V = γ P = B a . -V\frac{dP}{dV} = \gamma P = B_a. − V d V d P = γ P = B a .
3. Equation of Motion
Take a fluid slice with area A A A Δ x \Delta x Δ x
F net = − A ∂ p ∂ x Δ x . F_{\text{net}} = -A\frac{\partial p}{\partial x}\Delta x. F net = − A ∂ x ∂ p Δ x .
Its mass is m = ρ 0 A Δ x m = \rho_0 A\Delta x m = ρ 0 A Δ x
− A ∂ p ∂ x Δ x = ρ 0 A Δ x ∂ 2 η ∂ t 2 . -A\frac{\partial p}{\partial x}\Delta x = \rho_0 A\Delta x\frac{\partial^2\eta}{\partial t^2}. − A ∂ x ∂ p Δ x = ρ 0 A Δ x ∂ t 2 ∂ 2 η .
Cancel A Δ x A\Delta x A Δ x
− ∂ p ∂ x = ρ 0 ∂ 2 η ∂ t 2 . -\frac{\partial p}{\partial x} = \rho_0\frac{\partial^2\eta}{\partial t^2}. − ∂ x ∂ p = ρ 0 ∂ t 2 ∂ 2 η .
Use p = − B ∂ η ∂ x p = -B\frac{\partial \eta}{\partial x} p = − B ∂ x ∂ η
B ∂ 2 η ∂ x 2 = ρ 0 ∂ 2 η ∂ t 2 . B\frac{\partial^2\eta}{\partial x^2} = \rho_0\frac{\partial^2\eta}{\partial t^2}. B ∂ x 2 ∂ 2 η = ρ 0 ∂ t 2 ∂ 2 η .
So the 1D wave equation is
∂ 2 η ∂ x 2 = 1 c 2 ∂ 2 η ∂ t 2 . \boxed{\frac{\partial^2\eta}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2\eta}{\partial t^2}}. ∂ x 2 ∂ 2 η = c 2 1 ∂ t 2 ∂ 2 η .
General traveling wave form:
η ( x , t ) = f ( x − c t ) + g ( x + c t ) . \eta(x,t) = f(x-ct) + g(x+ct). η ( x , t ) = f ( x − c t ) + g ( x + c t ) .
Single frequency form:
η ( x , t ) = A cos ( k x − ω t + ϕ ) , ω = c k . \eta(x,t) = A\cos(kx-\omega t+\phi), \qquad \omega = ck. η ( x , t ) = A cos ( k x − ω t + ϕ ) , ω = c k .
Pressure perturbation:
p ( x , t ) = − B ∂ η ∂ x = B k A sin ( k x − ω t + ϕ ) . p(x,t) = -B\frac{\partial \eta}{\partial x} = BkA\sin(kx-\omega t+\phi). p ( x , t ) = − B ∂ x ∂ η = B k A sin ( k x − ω t + ϕ ) .
So pressure amplitude is
p m = B k A . p_m = BkA. p m = B k A .
Phasor form for a wave moving to + x +x + x
η ( x , t ) = η m e i ( ω t − k x ) . \eta(x,t) = \eta_m e^{i(\omega t-kx)}. η ( x , t ) = η m e i ( ω t − k x ) .
Derivatives:
η ˙ = i ω η , δ = ∂ η ∂ x = − i k η . \dot{\eta} = i\omega\eta, \qquad \delta = \frac{\partial \eta}{\partial x} = -ik\eta. η ˙ = iω η , δ = ∂ x ∂ η = − ik η .
Since s = − δ s=-\delta s = − δ
s = i k η . s = ik\eta. s = ik η .
With adiabatic modulus B a B_a B a
p = B a s = i B a k η . p = B_a s = iB_a k\eta. p = B a s = i B a k η .
5. Energy Density and Intensity
Particle speed is u = η ˙ u=\dot{\eta} u = η ˙
E k = 1 2 ρ 0 u 2 = 1 2 ρ 0 ( ∂ η ∂ t ) 2 . E_k = \frac{1}{2}\rho_0 u^2 = \frac{1}{2}\rho_0\left(\frac{\partial \eta}{\partial t}\right)^2. E k = 2 1 ρ 0 u 2 = 2 1 ρ 0 ( ∂ t ∂ η ) 2 .
With η = η m cos ( k x − ω t ) \eta=\eta_m\cos(kx-\omega t) η = η m cos ( k x − ω t )
E k = 1 2 ρ 0 ω 2 η m 2 sin 2 ( k x − ω t ) . E_k = \frac{1}{2}\rho_0\omega^2\eta_m^2\sin^2(kx-\omega t). E k = 2 1 ρ 0 ω 2 η m 2 sin 2 ( k x − ω t ) .
Time average:
⟨ E k ⟩ = 1 4 ρ 0 ω 2 η m 2 . \boxed{\langle E_k\rangle = \frac{1}{4}\rho_0\omega^2\eta_m^2}. ⟨ E k ⟩ = 4 1 ρ 0 ω 2 η m 2 .
Elastic potential energy density:
E p = 1 2 B s 2 = 1 2 B ( ∂ η ∂ x ) 2 . E_p = \frac{1}{2}Bs^2 = \frac{1}{2}B\left(\frac{\partial \eta}{\partial x}\right)^2. E p = 2 1 B s 2 = 2 1 B ( ∂ x ∂ η ) 2 .
For the same harmonic wave:
E p = 1 2 B k 2 η m 2 sin 2 ( k x − ω t ) . E_p = \frac{1}{2}Bk^2\eta_m^2\sin^2(kx-\omega t). E p = 2 1 B k 2 η m 2 sin 2 ( k x − ω t ) .
Since B = ρ 0 c 2 B=\rho_0c^2 B = ρ 0 c 2 ω = c k \omega=ck ω = c k
⟨ E p ⟩ = 1 4 ρ 0 ω 2 η m 2 . \boxed{\langle E_p\rangle = \frac{1}{4}\rho_0\omega^2\eta_m^2}. ⟨ E p ⟩ = 4 1 ρ 0 ω 2 η m 2 .
Total average energy density:
⟨ E ⟩ = 1 2 ρ 0 ω 2 η m 2 . \boxed{\langle E\rangle = \frac{1}{2}\rho_0\omega^2\eta_m^2}. ⟨ E ⟩ = 2 1 ρ 0 ω 2 η m 2 .
Using p m = ρ 0 c ω η m p_m=\rho_0 c\omega\eta_m p m = ρ 0 c ω η m
⟨ E ⟩ = p m 2 2 B = p m 2 2 ρ 0 c 2 . \langle E\rangle = \frac{p_m^2}{2B} = \frac{p_m^2}{2\rho_0 c^2}. ⟨ E ⟩ = 2 B p m 2 = 2 ρ 0 c 2 p m 2 .
Intensity for a forward plane wave:
I = ⟨ E ⟩ c = 1 2 ρ 0 c ω 2 η m 2 = p m 2 2 ρ 0 c . \boxed{I = \langle E\rangle c = \frac{1}{2}\rho_0 c\omega^2\eta_m^2 = \frac{p_m^2}{2\rho_0 c}}. I = ⟨ E ⟩ c = 2 1 ρ 0 c ω 2 η m 2 = 2 ρ 0 c p m 2 .
Part II, Sound Waves in Solids
1. Longitudinal Waves in a Rod
For a thin rod under uniaxial stress:
∂ 2 η ∂ x 2 = 1 c 2 ∂ 2 η ∂ t 2 , c 2 = Y ρ . \frac{\partial^2\eta}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2\eta}{\partial t^2}, \qquad c^2 = \frac{Y}{\rho}. ∂ x 2 ∂ 2 η = c 2 1 ∂ t 2 ∂ 2 η , c 2 = ρ Y .
Here Y Y Y
2. Isotropic 3D Elastic Constants
For isotropic solids, Poisson ratio is
σ = − ∂ β / ∂ y ∂ η / ∂ x . \sigma = -\frac{\partial \beta/\partial y}{\partial \eta/\partial x}. σ = − ∂ η / ∂ x ∂ β / ∂ y .
Useful identities:
σ = λ 2 ( λ + μ ) , λ = σ Y ( 1 + σ ) ( 1 − 2 σ ) , μ = Y 2 ( 1 + σ ) . \sigma = \frac{\lambda}{2(\lambda+\mu)}, \qquad
\lambda = \frac{\sigma Y}{(1+\sigma)(1-2\sigma)}, \qquad
\mu = \frac{Y}{2(1+\sigma)}. σ = 2 ( λ + μ ) λ , λ = ( 1 + σ ) ( 1 − 2 σ ) σ Y , μ = 2 ( 1 + σ ) Y . Bulk modulus:
B = λ + 2 3 μ = Y 3 ( 1 − 2 σ ) . B = \lambda + \frac{2}{3}\mu = \frac{Y}{3(1-2\sigma)}. B = λ + 3 2 μ = 3 ( 1 − 2 σ ) Y .
Wave speeds in bulk isotropic solids:
c L 2 = λ + 2 μ ρ = B + 4 3 μ ρ = Y ( 1 − σ ) ρ ( 1 + σ ) ( 1 − 2 σ ) , c_L^2 = \frac{\lambda+2\mu}{\rho}
= \frac{B+\frac{4}{3}\mu}{\rho}
= \frac{Y(1-\sigma)}{\rho(1+\sigma)(1-2\sigma)}, c L 2 = ρ λ + 2 μ = ρ B + 3 4 μ = ρ ( 1 + σ ) ( 1 − 2 σ ) Y ( 1 − σ ) , c T 2 = μ ρ = Y 2 ρ ( 1 + σ ) . c_T^2 = \frac{\mu}{\rho}
= \frac{Y}{2\rho(1+\sigma)}. c T 2 = ρ μ = 2 ρ ( 1 + σ ) Y . Typical Poisson ratio values:
Steel, σ ≈ 0.30 \sigma \approx 0.30 σ ≈ 0.30
Aluminum, σ ≈ 0.33 \sigma \approx 0.33 σ ≈ 0.33
Glass, σ ≈ 0.22 \sigma \approx 0.22 σ ≈ 0.22
Natural rubber, σ ≈ 0.49 \sigma \approx 0.49 σ ≈ 0.49
3. Shear Waves in Bulk Solids
Use transverse displacement β ( x , t ) \beta(x,t) β ( x , t )
T = μ ∂ β ∂ x . T = \mu\frac{\partial \beta}{\partial x}. T = μ ∂ x ∂ β .
Force balance on a thin slice gives
∂ ∂ x ( μ ∂ β ∂ x ) = ρ ∂ 2 β ∂ t 2 . \frac{\partial}{\partial x}\left(\mu\frac{\partial \beta}{\partial x}\right) = \rho\frac{\partial^2 \beta}{\partial t^2}. ∂ x ∂ ( μ ∂ x ∂ β ) = ρ ∂ t 2 ∂ 2 β .
For homogeneous material, μ \mu μ
μ ∂ 2 β ∂ x 2 = ρ ∂ 2 β ∂ t 2 . \mu\frac{\partial^2\beta}{\partial x^2} = \rho\frac{\partial^2\beta}{\partial t^2}. μ ∂ x 2 ∂ 2 β = ρ ∂ t 2 ∂ 2 β .
So
∂ 2 β ∂ x 2 = 1 c T 2 ∂ 2 β ∂ t 2 , c T 2 = μ ρ . \boxed{\frac{\partial^2\beta}{\partial x^2} = \frac{1}{c_T^2}\frac{\partial^2\beta}{\partial t^2}, \qquad c_T^2 = \frac{\mu}{\rho}}. ∂ x 2 ∂ 2 β = c T 2 1 ∂ t 2 ∂ 2 β , c T 2 = ρ μ .
Longitudinal bulk speed remains
c L 2 = λ + 2 μ ρ . c_L^2 = \frac{\lambda+2\mu}{\rho}. c L 2 = ρ λ + 2 μ .
This exceeds the thin rod estimate Y / ρ \sqrt{Y/\rho} Y / ρ 0 < σ < 1 / 2 0<\sigma<1/2 0 < σ < 1/2
Part III, Longitudinal Waves in Periodic Structures
1. Effective Stiffness and Young’s Modulus
Take a 1D atomic chain with spacing a a a K K K
Y = K a , K = Y a . \boxed{Y = \frac{K}{a}}, \qquad \boxed{K = Ya}. Y = a K , K = Y a .
With cell mass m ≈ ρ a 3 m\approx \rho a^3 m ≈ ρ a 3
ν ≈ 1 2 π K m ≈ 1 2 π a Y ρ ≈ c 0 2 π a . \nu \approx \frac{1}{2\pi}\sqrt{\frac{K}{m}}
\approx \frac{1}{2\pi a}\sqrt{\frac{Y}{\rho}}
\approx \frac{c_0}{2\pi a}. ν ≈ 2 π 1 m K ≈ 2 π a 1 ρ Y ≈ 2 π a c 0 . For a ≈ 2 × 10 − 10 m a\approx 2\times10^{-10}\,\text{m} a ≈ 2 × 1 0 − 10 m c 0 ≈ 5 × 10 3 m s − 1 c_0\approx 5\times10^3\,\text{m s}^{-1} c 0 ≈ 5 × 1 0 3 m s − 1
ν ≈ 3 × 10 12 Hz . \nu \approx 3\times10^{12}\,\text{Hz}. ν ≈ 3 × 1 0 12 Hz .
2. Discrete Model and Dispersion
Equation of motion for particle r r r
m η ¨ r = K ( η r + 1 + η r − 1 − 2 η r ) . m\ddot{\eta}_r = K(\eta_{r+1}+\eta_{r-1}-2\eta_r). m η ¨ r = K ( η r + 1 + η r − 1 − 2 η r ) .
Use trial form η r = η max e i ( ω t − k r a ) \eta_r=\eta_{\max}e^{i(\omega t-kra)} η r = η m a x e i ( ω t − k r a )
ω 2 = 4 K m sin 2 ( k a 2 ) . \boxed{\omega^2 = \frac{4K}{m}\sin^2\left(\frac{ka}{2}\right)}. ω 2 = m 4 K sin 2 ( 2 k a ) .
This result shows dispersion in a discrete medium.
Part IV, Reflection and Transmission at a Boundary
Take normal incidence at a boundary between two media.
Acoustic impedances are
Z 1 = ρ 1 c 1 , Z 2 = ρ 2 c 2 . Z_1=\rho_1 c_1, \qquad Z_2=\rho_2 c_2. Z 1 = ρ 1 c 1 , Z 2 = ρ 2 c 2 .
1. Boundary Conditions
At the interface:
η ˙ i + η ˙ r = η ˙ t , p i + p r = p t . \dot{\eta}_i+\dot{\eta}_r=\dot{\eta}_t, \qquad p_i+p_r=p_t. η ˙ i + η ˙ r = η ˙ t , p i + p r = p t .
For harmonic plane waves:
p = Z η ˙ for forward waves , p = − Z η ˙ for backward waves . p=Z\dot{\eta} \text{ for forward waves}, \qquad p=-Z\dot{\eta} \text{ for backward waves}. p = Z η ˙ for forward waves , p = − Z η ˙ for backward waves .
So:
p i = Z 1 η ˙ i , p r = − Z 1 η ˙ r , p t = Z 2 η ˙ t . p_i=Z_1\dot{\eta}_i, \qquad p_r=-Z_1\dot{\eta}_r, \qquad p_t=Z_2\dot{\eta}_t. p i = Z 1 η ˙ i , p r = − Z 1 η ˙ r , p t = Z 2 η ˙ t .
2. Velocity and Pressure Amplitude Coefficients
From continuity equations:
η ˙ r η ˙ i = Z 1 − Z 2 Z 1 + Z 2 , η ˙ t η ˙ i = 2 Z 1 Z 1 + Z 2 . \frac{\dot{\eta}_r}{\dot{\eta}_i}=\frac{Z_1-Z_2}{Z_1+Z_2}, \qquad \frac{\dot{\eta}_t}{\dot{\eta}_i}=\frac{2Z_1}{Z_1+Z_2}. η ˙ i η ˙ r = Z 1 + Z 2 Z 1 − Z 2 , η ˙ i η ˙ t = Z 1 + Z 2 2 Z 1 .
Pressure ratios:
p r p i = Z 2 − Z 1 Z 1 + Z 2 , p t p i = 2 Z 2 Z 1 + Z 2 . \frac{p_r}{p_i}=\frac{Z_2-Z_1}{Z_1+Z_2}, \qquad \frac{p_t}{p_i}=\frac{2Z_2}{Z_1+Z_2}. p i p r = Z 1 + Z 2 Z 2 − Z 1 , p i p t = Z 1 + Z 2 2 Z 2 .
If Z 1 > Z 2 Z_1>Z_2 Z 1 > Z 2 Z 1 < Z 2 Z_1<Z_2 Z 1 < Z 2
3. Intensity Coefficients
Use I = p r m s 2 / Z = Z ⟨ η ˙ 2 ⟩ r m s I=p_{\mathrm{rms}}^2/Z = Z\langle \dot{\eta}^2\rangle_{\mathrm{rms}} I = p rms 2 / Z = Z ⟨ η ˙ 2 ⟩ rms
Reflection and transmission fractions:
I r I i = ( Z 1 − Z 2 Z 1 + Z 2 ) 2 , I t I i = 4 Z 1 Z 2 ( Z 1 + Z 2 ) 2 . \boxed{\frac{I_r}{I_i}=\left(\frac{Z_1-Z_2}{Z_1+Z_2}\right)^2},
\qquad
\boxed{\frac{I_t}{I_i}=\frac{4Z_1Z_2}{(Z_1+Z_2)^2}}. I i I r = ( Z 1 + Z 2 Z 1 − Z 2 ) 2 , I i I t = ( Z 1 + Z 2 ) 2 4 Z 1 Z 2 . Energy check:
I r I i + I t I i = 1 . \boxed{\frac{I_r}{I_i}+\frac{I_t}{I_i}=1}. I i I r + I i I t = 1 .
Worked Problems
Problem 1, Open Open Pipe Pressure Mode
Given
∂ 2 p ∂ z 2 = ρ 0 B ∂ 2 p ∂ t 2 , \frac{\partial^2 p}{\partial z^2}=\frac{\rho_0}{B}\frac{\partial^2 p}{\partial t^2}, ∂ z 2 ∂ 2 p = B ρ 0 ∂ t 2 ∂ 2 p ,
with trial form
p ( z , t ) = [ C cos k z + D sin k z ] cos ω t , p(z,t)=\left[C\cos kz + D\sin kz\right]\cos\omega t, p ( z , t ) = [ C cos k z + D sin k z ] cos ω t ,
and condition p ( L / 2 , 0 ) = p 0 p(L/2,0)=p_0 p ( L /2 , 0 ) = p 0
Open ends set pressure perturbation to zero:
p ( 0 , t ) = 0 ⇒ C = 0 , p(0,t)=0 \Rightarrow C=0, p ( 0 , t ) = 0 ⇒ C = 0 ,
p ( L , t ) = 0 ⇒ sin ( k L ) = 0 ⇒ k = n π L . p(L,t)=0 \Rightarrow \sin(kL)=0 \Rightarrow k=\frac{n\pi}{L}. p ( L , t ) = 0 ⇒ sin ( k L ) = 0 ⇒ k = L nπ .
At z = L / 2 z=L/2 z = L /2 t = 0 t=0 t = 0
p 0 = D sin ( n π 2 ) . p_0=D\sin\left(\frac{n\pi}{2}\right). p 0 = D sin ( 2 nπ ) .
So only odd n n n
D = p 0 sin ( n π / 2 ) = ( − 1 ) n − 1 2 p 0 . D=\frac{p_0}{\sin(n\pi/2)} = (-1)^{\frac{n-1}{2}}p_0. D = s i n ( nπ /2 ) p 0 = ( − 1 ) 2 n − 1 p 0 .
Substitute into wave equation:
k 2 = ρ 0 ω 2 B . k^2=\frac{\rho_0\omega^2}{B}. k 2 = B ρ 0 ω 2 .
Then
ω = n π L B ρ 0 , B = ρ 0 ( ω L n π ) 2 , n = 1 , 3 , 5 , … \omega=\frac{n\pi}{L}\sqrt{\frac{B}{\rho_0}}, \qquad B=\rho_0\left(\frac{\omega L}{n\pi}\right)^2, \qquad n=1,3,5,\dots ω = L nπ ρ 0 B , B = ρ 0 ( nπ ω L ) 2 , n = 1 , 3 , 5 , …
Practical note.
Boundary and initial data fix mode index and amplitude.
You still need one material or frequency datum to fix both B B B ω \omega ω
Problem 2, Thermal Speed Versus Sound Speed
From kinetic theory:
1 2 m v r m s 2 = 3 2 k B T ⇒ v r m s 2 = 3 k B T m . \frac{1}{2}mv_{\mathrm{rms}}^2=\frac{3}{2}k_B T \Rightarrow v_{\mathrm{rms}}^2=\frac{3k_B T}{m}. 2 1 m v rms 2 = 2 3 k B T ⇒ v rms 2 = m 3 k B T .
For ideal gas sound:
c 2 = γ P ρ = γ k B T m . c^2=\frac{\gamma P}{\rho}=\frac{\gamma k_B T}{m}. c 2 = ρ γ P = m γ k B T .
Take the ratio:
v r m s c = 3 γ . \frac{v_{\mathrm{rms}}}{c}=\sqrt{\frac{3}{\gamma}}. c v rms = γ 3 .
For monoatomic gas, γ = 5 / 3 \gamma=5/3 γ = 5/3 ≈ 1.34 \approx1.34 ≈ 1.34 γ = 7 / 5 \gamma=7/5 γ = 7/5 ≈ 1.46 \approx1.46 ≈ 1.46
Problem 3, Acoustic Pressure at Pain Threshold
Given ρ = 1.29 k g / m 3 \rho=1.29\,\mathrm{kg/m^3} ρ = 1.29 kg/ m 3 c = 330 m / s c=330\,\mathrm{m/s} c = 330 m/s I = 10 W / m 2 I=10\,\mathrm{W/m^2} I = 10 W/ m 2
I = p r m s 2 ρ c ⇒ p r m s = I ρ c . I=\frac{p_{\mathrm{rms}}^2}{\rho c} \Rightarrow p_{\mathrm{rms}}=\sqrt{I\rho c}. I = ρ c p rms 2 ⇒ p rms = I ρ c .
Compute:
p r m s = 10 × 1.29 × 330 ≈ 65 P a . p_{\mathrm{rms}}=\sqrt{10\times1.29\times330}\approx65\,\mathrm{Pa}. p rms = 10 × 1.29 × 330 ≈ 65 Pa .
Convert to atmosphere:
p r m s 1 a t m = 65 1.013 × 10 5 ≈ 6.4 × 10 − 4 . \frac{p_{\mathrm{rms}}}{1\,\mathrm{atm}}=\frac{65}{1.013\times10^5}\approx6.4\times10^{-4}. 1 atm p rms = 1.013 × 1 0 5 65 ≈ 6.4 × 1 0 − 4 .
Problem 4, Displacement Amplitude at Pain Threshold, 500 Hz
Use plane wave relation:
I = 1 2 ρ 0 c ω 2 η m 2 , ω = 2 π f . I=\frac{1}{2}\rho_0 c\,\omega^2\eta_m^2, \qquad \omega=2\pi f. I = 2 1 ρ 0 c ω 2 η m 2 , ω = 2 π f .
So
η m = 1 2 π f 2 I ρ 0 c . \eta_m=\frac{1}{2\pi f}\sqrt{\frac{2I}{\rho_0 c}}. η m = 2 π f 1 ρ 0 c 2 I .
With I = 10 W / m 2 I=10\,\mathrm{W/m^2} I = 10 W/ m 2 ρ 0 = 1.29 k g / m 3 \rho_0=1.29\,\mathrm{kg/m^3} ρ 0 = 1.29 kg/ m 3 c = 330 m / s c=330\,\mathrm{m/s} c = 330 m/s f = 500 H z f=500\,\mathrm{Hz} f = 500 Hz
η m ≈ 6.9 × 10 − 5 m . \eta_m\approx6.9\times10^{-5}\,\mathrm{m}. η m ≈ 6.9 × 1 0 − 5 m .
Problem 5, Near Inaudible Tone at 500 Hz
Take I = 10 − 10 W / m 2 I=10^{-10}\,\mathrm{W/m^2} I = 1 0 − 10 W/ m 2
η m = 1 2 π f 2 I ρ 0 c . \eta_m=\frac{1}{2\pi f}\sqrt{\frac{2I}{\rho_0 c}}. η m = 2 π f 1 ρ 0 c 2 I .
With f = 500 H z f=500\,\mathrm{Hz} f = 500 Hz ρ 0 = 1.29 k g / m 3 \rho_0=1.29\,\mathrm{kg/m^3} ρ 0 = 1.29 kg/ m 3 c = 330 m / s c=330\,\mathrm{m/s} c = 330 m/s
η m ≈ 2.2 × 10 − 10 m . \eta_m\approx2.2\times10^{-10}\,\mathrm{m}. η m ≈ 2.2 × 1 0 − 10 m .
This lies on the molecular size scale.
Reference
H. J. Pain, The Physics of Vibrations and Waves, 6th ed., Wiley, 2013.
