Sound Waves: Propagation, Velocity, and Intensity

Posted on Jan 15, 2025 in Physics

Introduction to Sound Waves

Part of the study of physics is the movement of sound. Sound waves, or pressure waves, can only propagate through material media by varying the relative pressure of the medium. Sound, like every wave, has characteristics such as amplitude, frequency, wavelength, and propagation speed. The variations in pressure values compared with any sound wave are very small (measured in dB). The frequency of sound establishes a classification: 20-20,000 Hz is audible sound, below 20 Hz is infrasound, and above 20,000 Hz is ultrasound. The propagation velocity depends on the mechanical characteristics of the medium. Taking these values into account, we can deduce the corresponding wavelength: λ = c / f.

Shear Wave Velocity

For an inextensible string, the resting force at its extremes is the same. When a transverse wave propagates along the x-axis, the elements move on the y-axis in the transverse direction. We can express this as dF_y = T₀(sin θ₂ – sin θ₁). Assuming a small angle, we can approximate sin θ with tan θ, so dF_y = T₀d(tan θ). Multiplying by the length (dx), we get (dF_y / dx) = T₀[d(tan θ) / dx] = T₀(d²y / dx²). Applying Newton’s second law, we have ρ = m / dx. Comparing this with the equation of propagation, we find the transverse velocity c_t = √(T₀ / ρ). The equation for a transverse plane wave propagating along a string is y = Asin[ω(t – (x / c_t))].

Longitudinal Wave Velocity

For a circular section rod made of an elastic material, when a wave passes, it causes an elastic deformation. We can apply Hooke’s law here. The passage of this wave undergoes a tensile stress proportional to the deformation: (dF_x / A) = E[(L_f – L₀) / L₀]. The force can be calculated as dF_x = AE(ΔL / L₀). Assuming an extreme range with amplitude du, we have (ΔL / L₀) = (L_f – L₀) / L₀ = (du / dx). Substituting this into the previous equation, we get (dF / dx) = AE(d²u / dx²). Applying Newton’s second law, dF / dx = (ρAdx / dx)(d²u / dt²). Comparing this with the wave equation, we find the longitudinal velocity in a solid c_ls = √(E / ρ). The wave equation is u = Asin[ω(t – (x / c_ls))].

Fluid

For fluids, c_Lf = √(1 / ρk). The wave equation is u = Asin[ω(t – (x / c_Lf))], where u represents the wave amplitude.

Pressure Waves

Sound waves correspond to pressure waves. If we consider an incident wave and a reflected wave with the same amplitude, their equations are: dp_i = Asin[2πf(t – (x / c))] and dp_r = Asin[2πf(t + (x / c))]. The resulting wave is the sum of these two: dp = 2Acos(2πfx / c)sin(2πft). The amplitude is A_R = 2Acos(2πx / λ), and the maximum pressure points are x_Vp = k(λ / 2), while the pressure nodes are x_Np = (2k + 1)(λ / 4). To obtain the particle displacement, we start with the wave equation dp = -(1 / k)(du / dx) = -k2Acos(2πx / λ)sin(2πft)dx. Integrating, we get u = -(2Ak / 2π)sin(2πx / λ)sin(2πft). The velocity of each particle is u = du / dt = -2Akωfsin(2πx / λ)cos(2πft).

Bernoulli’s Theory of Sound in Tubes

A tube is a sound source capable of producing sounds usable in music. The tube’s axis is x, with the origin where the primary wave is introduced. This produces a standing wave, which is the superposition of two waves of equal frequency propagating in opposite directions along the x-axis: u₁ = A₁sin[2πf(t – (x / v))] and u₂ = A₂sin[2πf(t + (x / v))]. The resulting wave is the sum of these two: U = A₁sin[2πf(t – (x / v))] + A₂sin[2πf(t + (x / v))]. Considering that the pressure must be the same outside, (du / dx)_x=0 = (2πf / c)(-A₁cos[2πf(t – (x / c))] + A₂cos[2πf(t + (x / c))]) = 0, which implies A₁ = A₂. Therefore, u = 2Acos(2πfx / c)sin(2πft).

Open Tubes

It is verified that dp = -(1 / k)(du / dx)_x=La = 0. Using the equation for dp, we get dp = -(1 / k)(du / dx) = 2A(2πf / kc)sin(πfx / c)sin(2πft) = 0. This implies sin(2πfL_a / c) = 0, indicating that the frequency as a function of the tube length is L_a = c / 2f = λ / 2. An open tube emits a frequency f = kc / 2L_a.

Closed Tubes

In a tube closed at one end, the particles do not vibrate (u)_x=Lc = 0. We obtain a wave equation 2πfL_c / c = (2k – 1)(π / 2). From this, we deduce the function of the tube length and frequency: L_c = (2k – 1)(λ / 4). A closed tube has a frequency f_c = [(2k – 1)c] / 4L_c.

Sound Intensity

Sound intensity is the energy that crosses per second a unit surface normal to the propagation direction. It is given by I = A_p² / (ρc). The sound level is given by ss₀ = log(I / I₀).

Doppler Effect

The Doppler effect is the variation in frequency that an observer perceives when an oscillatory focus and the observer are in motion.