Part A: Introduction to Acoustics
Part II. Review of Driven-Damped Oscillations
Consider a 1 DoF (degree of freedom) system consisting of a spring and mass damper. The spring has a stiffness of s and the damper has a damping coefficient of R_m. A loading force F(t) is applied to the mass, as shown in the image below.
From Newton’s 2nd law, the force balance for the system is
m\ddot x+R_m\dot x+sx=F(t) \tag{1}
The displacement solution x(t) can be solved for and expressed as the sum of a transient solution and a steady-state solution. The transient solution is dependent on time and naturally decays to zero as t\to\infty. The steady-state solution is a sinusoidal function that has a frequency of \omega.
Steady-State Solution
The general form for the steady-state solution is
x(t)=A\cos(\omega t+\phi)
where A is the amplitude of the wave, \omega is the frequency as discussed above, and \phi is the phase angle or phase lag. The phase lag is responsible for the delay between when the loading force is applied and when the system registers this force and affects the mass’ position. To simplify the arithematic, we may introduce the complex unit j to describe harmonic oscillation. For instance, the loading force F(t) may be expressed as the real component of the complex loading force \underline F(t).
\underline F(t)=F_0e^{j\omega t}=F_0\cos\omega t+F_0j\sin\omega t \tag{2}
And obviously, the physical loading force is the real part of Eq. 2.
F(t)=\mathrm{Re}\left[\underline F(t)\right]=F_0\cos\omega t
Similarly, the displacement x(t) can be thought of as the real component of the complex displacement \underline x(t).
\underline x(t)=Ae^{j(\omega t+\phi)}=\left(Ae^{j\phi}\right)e^{j\omega t}=\underline Ae^{j\omega t} \tag{3}
The coefficient of \exp(j\omega t) in Eq. 3 is called the complex amplitude \underline A.
Introducing complex numbers may seem like an odd choice initially. However, as we will see throughout the course, they simplify the math when deriving these equations of motion. The derivative of \exp(j\omega t) is much easier to compute than the raw trigonometric functions themselves. Additionally, because of Euler’s formula
\left(e^{j\omega t}\right)^n=e^{j\omega nt}=\cos\omega nt+j\sin\omega nt
raising complex numbers to some power becomes much simpler to handle.
We may substitute the polar form of the complex numbers into Eq. 1 to derive a simpler form. The first and second order derivatives of \underline x(t) are
\begin{aligned} \dot{\underline x} & =j\omega\underline x \\ \ddot{\underline x} & =j\omega\dot{\underline x}=-\omega^2\underline x \end{aligned}
As a result, Eq. 1 becomes
F_0e^{j\omega t}=(-m\omega^2+j\omega R_m+s)\underline x=(-m\omega^2+j\omega R_m+s)\underline Ae^{j\omega t}
Cancelling like-terms and solving for the complex amplitude \underline A gives
\underline A=\frac {F_0}{s-m\omega^2+j\omega R_m} \tag{4}
The magnitude \left|\underline A\right| and polar angle \phi are simply
\begin{aligned} \left|\underline A\right| & =\frac {F_0}{\sqrt{(s-m\omega^2)^2+(\omega R_m)^2}} \\ \phi & =-\arctan\left(\frac {\omega R_m}{s-m\omega^2}\right) \end{aligned} \tag{5}
Mechanical Impedance
Physical Velocity
Resonance Frequency
Power Consumption
When talking about waves, it is important to discuss the power associated with a specific waveform. The instantaneous power is denoted by \Pi_i(t) and represents the instantaneous power carried by the wave. This quantity is a real, physical, time-dependent value. The power can be computed based off the wave’s force F(t) and propagation velocity u(t).
\Pi_i(t)=F(t)u(t)=\mathrm{Re}\left[\underline F(t)\right]\mathrm{Re}\left[\underline u(t)\right] \tag{6}
Observe that the \mathrm{Re}(\cdot) function is applied to each term separately. That is, the real part of \underline F(t) and \underline u(t) are carried out in separate functions. This is because \mathrm{Re}(\cdot) cannot be “distributed”.
\mathrm{Re}\left[\underline F(t)\underline u(t)\right]\neq\mathrm{Re}\left[\underline F(t)\right]\mathrm{Re}\left[\underline u(t)\right]=F(t)u(t)
Typically, it is useful to compute the time-averaged power \Pi_{\mathrm{avg}}. This is simply calculated by integral averaging the instantaneous power across a suitable averaging time interval T_{\mathrm{avg}}. For harmonics, we use the period: T_{\mathrm{avg}}=T=2\pi/\omega.
\Pi_{\mathrm{avg}}=\frac 1T\int\limits_0^T\Pi_i(t)\,\mathrm dt=\frac {F_0^2}{2\left|\underline z_m\right|}\cos\theta_z
where \underline z_m=R_m+X_mj is the complex mechanical impedance and \theta_z is the phase angle associated with \underline z_m. Here, it can be shown that \cos\theta_z=R_m/\left|\underline z_m\right| so
\Pi_{\mathrm{avg}}=\frac {F_0^2R_m}{2\left|\underline z_m\right|^2}=\frac {F_0^2R_m}{2\left[R_m^2+\left(\omega m-\frac sm\right)^2\right]} \tag{7}
Part III. Wave Propagation and Vibration of a 1D System
Consider a string of length L vibrating between two solid walls. The string is fixed at both ends, x=0 and x=L, and let y(x, t) be the transverse displacement of a point on the string. The equation of motion for the string is described by the wave equation.
\frac {\partial^2y}{\partial x^2}=\frac 1{c^2}\frac {\partial^2 y}{\partial t^2} \tag{8}
where c=\sqrt{T/\rho_L} is the wave speed, T is the tension in the string, and \rho_L is the linear density of the string. In SI units, \rho_L will have units of kg/m. It is worth pointing out, however, that Eq. 8 makes a series of underlying assumptions.
The string undergoes small deflection, defined as a small maximum transverse displacement relative to its length. Mathematically, this means y_{\mathrm{max}}/L\ll1 or \partial y/\partial x\ll1.
The tension and string linear density are constant and uniform throughout the length of the string.
The effects of gravity are negligible.
No damping is present in the string.
The general solution is given by D’Alembert’s formula
y(x, t)=y_1(ct-x)+y_2(ct+x) \tag{9}
where y_1 and y_2 are two arbitrary functions.
In Eq. 9, y_1 and y_2 are unknown functions that depend on the boundary conditions and initial conditions.
To illustrate the procedure for solving for y_1 and y_2, consider a string of infinite length subject to the initial conditions below. Note that two are needed since the PDE is 2nd order.
\begin{aligned} y(x, 0) & =y_0(x) \\ \frac {\partial y}{\partial y}(x, 0) & =0 \end{aligned}
We may substitute into Eq. 9 to obtain a system of equation.
\begin{aligned} y_1(-x)+y_2(x) & =y_0(x) \\ cy_1'(-x)+cy_2'(x) & =0 \end{aligned}
The second equation in y'(\pm x) above implies after integrating that -y_1(-x)+y_2(x)=\alpha for an arbitrary constant \alpha. Therefore, y_2(x)=y_1(-x)+\alpha. Substitute this equation for y_2 into the first equation and solve for y_1(x) to arrive at
\begin{aligned} y_1(x) & =\frac 12\left[y_0(-x)+\alpha\right] \\ y_2(x) & =\frac 12\left[y_0(x)+\alpha\right] \end{aligned}
Since \alpha is any constant, we may assume \alpha=0 without loss of generality so the solution becomes
y(x, t)=\frac 12\left[y_0(x-ct)+y_0(x+ct)\right]
There is a physical meaning to y(x, t) above. The y_0(x-ct) term represents a wave propagating to the right and y_0(x+ct) represents an identical wave propagating to the left. Each wave propagates at a speed set by the wave speed, c.
Let’s consider the example of a string of length L with both ends fixed. The string is subject to the initial conditions
\begin{aligned} y(x, 0) & =y_0(x) \\ \frac {\partial y}{\partial t}(x, 0) & =0 \end{aligned} \tag{10}
Since Eq. 8 has no damping, we note that the motion will be periodic as the waves travel through the length of the string without losing energy. Therefore, we may assume that
y_m(x, t)=\left[A_m\sin k_mx+B_m\cos k_mx\right]e^{j\omega_m t}
Since the string cannot move at the ends due to the fixed boundary conditions, then
\begin{aligned} y(0, t) & =B_me^{j\omega_m t}=0 \\ y(L, t) & =A_m\sin k_mLe^{j\omega_m t}=0 \end{aligned}
From the first equation, B_m=0 since \exp(j\omega_m t)>0. With the second equation, there is indeed the case that A_m=0. However, this reduces y(x, t) down to the trivial case as A_m=B_m=0 means there is no vibration. Therefore, we must consider the non-trivial case where \sin k_mL=0. This occurs at unit intervals of k_m=n\pi/L called the wave number.
Now, we consider specific values of n. When n=0, k_0=0 and y_0(x, t) again reduces down to the trivial case. The result is
\omega_n=\frac {n\pi c}L \tag{11}
Eq. 11 is called the natural frequency of a standing wave or of a node n. Using \sin x=(e^{jx}-e^{-jx})/(2i) gives
The total free vibration response is simply the sum of each vibration response.
y(x, t)=\sum\limits_{m=1}^{+\infty}y_m(x, t)=\sum\limits_{m=1}^{+\infty}A_m\sin\left(\frac {n\pi x}L\right)e^{j\omega_n t} \tag{12}
The value for A_m depends on the initial condition. Substitute the initial conditions into Eq. 12 to compute A_m.
Each n in Eq. 12 describes a specific mode shape. For n=1, y_1(x)=\sin\left(\frac {\pi x}L\right). This represents a string vibrating with the greatest displacement at x=\frac L2. When n=2, this is the harmonic as y_2(x)=\sin\left(\frac {2\pi x}L\right)=\sin 2\omega_1 where \omega_1 is the natural frequency (when n=1). The wave shape for this mode is that of a string vibrating where the center x=\frac L2 does not move.