3. Oscillations and WavesSavchenko Formulas, chapter 3 of 14, 34 formulas
Sections follow the book, and within a section the most used formulas come first. An italic problem number means the formula appears in its statement. Rest the cursor on a number to see the statement.
$\displaystyle \varkappa = U''(x_0)$$\displaystyle T = 2\pi\sqrt{\frac{m}{U''(x_0)}}$
x
displacement from equilibrium
$\varkappa$
effective stiffness, the second derivative of the potential energy
m
mass
Near a minimum of potential energy the curve is a parabola, and any system oscillates harmonically at $\sqrt{\varkappa/m}$. The restoring force follows from expanding the force in the small displacement and keeping the linear term, as $\sin\varphi \approx \varphi$ for a pendulum.
$$\omega = \sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}}$$
$\displaystyle T = 2\pi\sqrt{\frac{\Delta l}{g}}\ \text{(}\Delta l\text{ статическое растяжение)}$$\displaystyle \omega = \sqrt{\frac{k}{\mu}},\quad \mu = \frac{m_1 m_2}{m_1 + m_2}\ \text{(два тела на пружине)}$
k
spring constant
m
mass of the load
$\Delta l$
static extension under the load
The equation $m\ddot x = -kx$ gives harmonic motion at $\sqrt{k/m}$, independent of the amplitude. Gravity only shifts the equilibrium. For two bodies on one spring the mass is replaced by the reduced mass.
$\displaystyle T = 2\pi\sqrt{\frac{l}{g\cos\alpha}}\ \text{(на наклонной плоскости)}$$\displaystyle \omega = \sqrt{\frac{g}{R}}\ \text{(шарик в сферической чаше, тело в туннеле сквозь Землю)}$
l
length of the string
g
free-fall acceleration or its effective component
For small angles $\sin\varphi \approx \varphi$ and the restoring force is $mg\,x/l$, hence $\omega^2 = g/l$. The period depends neither on the mass nor on the amplitude. A bead sliding in a bowl of radius $R$ and a body in a tunnel through a uniform Earth oscillate at $\omega = \sqrt{g/R}$.
coefficient of the squared coordinate in the potential energy
$\beta$
coefficient of the squared velocity in the kinetic energy
Writing the system's total energy through one coordinate and its rate, and keeping the quadratic terms, gives the frequency as the square root of the ratio of the coefficients, without writing any forces. This suits systems with pulleys, liquid in a tube, rods and springs together.
In an accelerating frame the inertial force $-m\vec a$ adds to gravity, and the pendulum swings about a new equilibrium with period $2\pi\sqrt{l/g^*}$. In a freely falling lift $g^* = 0$ and there is no oscillation. The same $g^*$ sets the period at another altitude or on another planet.
$\displaystyle T = 2\pi\sqrt{\frac{2l}{3g}}\ \text{(стержень за конец)}$$\displaystyle T = 2\pi\sqrt{\frac{2R}{g}}\ \text{(кольцо на гвозде)}$
I
moment of inertia about the pivot
a
distance from the pivot to the centre of mass
m
mass
A body swinging about an axis obeys $I\ddot\varphi = -mga\sin\varphi$, harmonic for small angles. The equivalent length $I/ma$ is the length of a simple pendulum with the same period. The moment of inertia comes from the parallel axis theorem.
$$T = 2\pi\sqrt{\frac{H}{g}}, \qquad T = 2\pi\sqrt{\frac{m}{\rho g S}}$$
$\displaystyle \omega = \sqrt{\frac{2g}{l}}\ \text{(столб жидкости длины }l\text{ в U-трубке)}$
H
draught of the floating body or half the column length
S
cross-section
$\rho$
density of the liquid
Displacing a floating body by $x$ changes the buoyant force by $\rho g S x$, a restoring force of stiffness $\rho g S$, and the body's mass is $\rho S H$, hence $\omega^2 = g/H$. For a liquid column in a tube the level difference $2x$ gives a force $2\rho g S x$ on the whole column.
In the centre-of-mass frame the two bodies move towards and away from each other, and their relative motion is that of one body of the reduced mass. The centre of mass moves uniformly. The vibration frequency of a diatomic molecule is found the same way.
$$x = A\cos(\omega t + \varphi_0), \qquad v = -A\omega\sin(\omega t + \varphi_0), \qquad a = -\omega^2 x$$
$\displaystyle v_{\max} = A\omega,\quad a_{\max} = A\omega^2$$\displaystyle x = A\sin\omega t\ \text{(из положения равновесия)}$
A
amplitude
$\omega$
angular frequency, $2\pi/T$
$\varphi_0$
initial phase
The coordinate follows a sine or cosine, the velocity leads it by a quarter period, the acceleration is opposite to the coordinate. The phase and amplitude follow from the position and velocity at the start. The projection of uniform circular motion on a diameter is harmonic motion.
$\displaystyle m\ddot x = -kx$$\displaystyle \ddot\varphi + \frac{g}{l}\varphi = 0$
x
displacement from equilibrium
$\omega$
angular frequency
Every equation of the form $\ddot x = -\omega^2 x$ is solved by $A\cos(\omega t + \varphi_0)$, so once Newton's second law is brought to this form the frequency is read off as the square root of the coefficient of $x$. A constant force on the right only shifts the equilibrium and leaves the frequency unchanged.
$$t = \frac{\Delta\varphi}{\omega}, \qquad T = T_1 + T_2$$
$\displaystyle \sin\omega t = \frac{1}{2}\ \Rightarrow\ t = \frac{T}{12}$$\displaystyle T = \pi\sqrt{\frac{l_1}{g}} + \pi\sqrt{\frac{l_2}{g}}\ \text{(маятник с гвоздём)}$
$\Delta\varphi$
phase change between two positions
T1, T2
periods of the motion on the two parts
The time between two positions comes from the phase, not from the path, since the speed is not constant. From equilibrium to half the amplitude takes $T/12$, to the amplitude $T/4$. When a pendulum changes its length or stiffness halfway, the period is the sum of the halves of the two periods.
The total energy of an oscillator is proportional to the amplitude squared and passes from kinetic to potential twice a period. Averaged over a period they are equal. The amplitude for a given position and velocity follows from energy conservation.
The general solution of the oscillator equation is a cosine plus a sine with coefficients from the initial conditions. A suddenly applied constant force shifts the equilibrium by $F/k$, and the body oscillates about it with amplitude $F/k$, reaching $2F/k$, twice the static extension.
$$\Delta A = \frac{2\mu mg}{k}\ \text{за полпериода}$$
$\displaystyle \omega = \sqrt{\frac{2\mu g}{l}}\ \text{(доска на двух вращающихся валиках)}$
$\mu$
coefficient of friction
k
stiffness
$\Delta A$
loss of amplitude every half period
Dry friction is a constant force that flips with the velocity, so every half period the body oscillates harmonically about an equilibrium shifted by $\mu mg/k$, and the amplitude falls arithmetically by $2\mu mg/k$. Motion stops when the amplitude drops below $\mu mg/k$.
A body on a platform presses on it with $m(g - a)$,$a$ being the platform's upward acceleration. It leaves when the downward acceleration reaches $g$, that is when $A\omega^2 \ge g$, at the point where $\omega^2 y = g$. It then flies freely until it meets the platform again.
$\displaystyle x = x_1 + x_2\ \text{(принцип суперпозиции)}$$\displaystyle A = 2A_1\cos\frac{\Delta\varphi}{2}\ \text{(равные амплитуды)}$
A1, A2
amplitudes of the two oscillations
$\varphi_2 - \varphi_1$
phase difference
Two oscillations of one frequency along one line add to an oscillation of the same frequency, the amplitude by the vector rule with the phase difference as the angle. In phase the amplitudes add, in antiphase they subtract. Waves and pulses add the same way, point by point.
beat period, the time between two amplitude maxima
The sum of two close frequencies is an oscillation at the mean frequency with a slowly varying amplitude that vanishes whenever the two drift $\pi$ apart in phase. The number of beats per second equals the frequency difference. An undamped oscillator driven near resonance beats too.
numbers of tangencies with the sides of the bounding rectangle
A point oscillating along two perpendicular axes at one frequency moves on an ellipse, on a straight segment when the phases differ by $0$ or $\pi$, on a circle at $\pi/2$ with equal amplitudes. Commensurate frequencies give Lissajous figures, and the frequency ratio equals the ratio of the numbers of tangencies with the horizontal and vertical sides.
$\displaystyle x_1 = B\cos(\omega_1 t + \varphi) + A\cos\omega_2 t,\quad x_2 = B\cos(\omega_1 t + \varphi) - A\cos\omega_2 t$
$\omega_0$
frequency of each pendulum alone
k
stiffness of the coupling spring
$\omega_1, \omega_2$
frequencies of the in-phase and antiphase modes
Two identical coupled pendulums have two motions of a single frequency, the in-phase one where the spring is idle and the antiphase one where it is stretched doubly. Any motion is their sum with coefficients from the initial conditions, and energy passes from one pendulum to the other at the beat frequency $\omega_2 - \omega_1$.
$\displaystyle A_{\text{рез}} = \frac{F_0}{2m\gamma\omega_0} = Q\frac{F_0}{k}$$\displaystyle \omega \ll \omega_0:\ x \approx \frac{F}{k},\qquad \omega \gg \omega_0:\ x \approx -\frac{F}{m\omega^2}$
F0
amplitude of the driving force
$\omega$
its frequency
$\omega_0$
natural frequency
$\gamma$
damping constant
$\varphi$
phase lag of the displacement behind the force
The steady oscillation runs at the driving frequency, its amplitude is largest near the natural frequency and for weak damping exceeds the static displacement $F_0/k$ by the factor $Q$. A slow force is followed, a fast one is lagged by $\pi$ and barely moves the body. Without damping the amplitude at resonance grows linearly in time.
Viscous friction proportional to velocity makes the amplitude decay exponentially and slightly lowers the frequency. The ratio of amplitudes one period apart is constant. The quality factor tells how many periods over $2\pi$ the oscillation lives, and equals the stored energy over the loss per radian.
$$\Delta v = \frac{p_0}{m}, \qquad p_n = m v_0 + n p_0$$
p0
impulse of one kick
m
mass
$\omega$
natural frequency
A short kick changes the velocity by $p_0/m$ without moving the body. Kicks in step with the motion add up and drive it, kicks in antiphase damp it. With friction the steady amplitude follows from equating a kick's energy to the loss per period.
$$\Delta p = \rho c\, u, \qquad \varepsilon = -\frac{u}{c}$$
$\displaystyle \Delta p_{\max} = \rho c\,\omega A$$\displaystyle F = \rho c S u\ \text{(сила на торец стержня)}$
$\Delta p$
excess pressure or stress in the wave
u
velocity of the medium's particles, not of the wave
$\rho c$
wave impedance of the medium
In a time $dt$ the wave sets in motion a layer of thickness $c\,dt$ and mass $\rho c S\,dt$, giving it the velocity $u$, hence the force $\rho c S u$ and pressure $\rho c u$. The strain in the wave is $u/c$. The product $\rho c$ governs how a wave reflects at the boundary of two media.
$\displaystyle \Delta L = \frac{FL}{ES}$$\displaystyle \sigma_{\text{терм}} = E\alpha\Delta T\ \text{(зажатый стержень при нагреве)}$
$\sigma$
stress, force per unit cross-section
$\varepsilon$
strain
E
Young's modulus
S, L
cross-section and length of the rod
Stress is proportional to strain, the coefficient being Young's modulus, a property of the material rather than the sample. A rod behaves as a spring of stiffness $ES/L$. Under its own weight or acceleration the stress varies along the length, and the extension is an integral over elements.
$\displaystyle c = \sqrt{\frac{\gamma RT}{\mu}}\ \text{(звук в газе)}$$\displaystyle c = \sqrt{gh}\ \text{(волны на мелкой воде)}$
E
Young's modulus
$\rho$
density of the medium
F
tension of the string
$\rho_l$
mass per unit length of the string
$\gamma$
adiabatic index of the gas
A wave's speed is the square root of the medium's elasticity over its inertia, the density. In a rod that is $E/\rho$, in a string the tension over the mass per length, in a gas the adiabatic elasticity $\gamma p$ over the density, giving $\sqrt{\gamma RT/\mu}$. The speed depends neither on frequency nor on amplitude.
A stretched rod narrows, and the ratio of the transverse to the longitudinal strain is Poisson's ratio, about $0.3$ for metals and $0.5$ for rubber, where the volume stays fixed. Compressibility relates to it through $\kappa = 3(1 - 2\nu)/E$.
As a spring stores $kx^2/2$, a unit volume of strained body stores $E\varepsilon^2/2$. In a travelling elastic wave the kinetic energy density equals the potential one at every point.
wavelength, the distance between neighbouring crests
k
wave number
c
wave speed
$\nu, T$
frequency and period of the oscillation at each point
Any function of $x - ct$ is a wave running to the right at speed $c$ without change of shape. In one period the wave travels one wavelength, so frequency, wavelength and speed obey $c = \lambda\nu$. A point at a distance $x$ oscillates with a delay $x/c$.
angles between the ray and the normal to the boundary
c1, c2
wave speeds in the two media
v
speed of the source, faster than the wave
In a time $\Delta t$ the wavefront advances $c_1\Delta t$ in one medium and $c_2\Delta t$ in the other, and Huygens' construction gives the law of refraction. Where the speed varies smoothly, $c/\sin\theta$ is constant along the ray. A source faster than the wave makes a cone with $\sin\alpha = c/v$.
$$p = \rho c\, u S\tau = F\tau, \qquad W = \rho c S u^2 \tau$$
$\tau$
duration of the pulse
u
particle velocity in the pulse
F
force at the end that made the pulse
A pulse of duration $\tau$ occupies a stretch of length $c\tau$ and mass $\rho S c\tau$ moving at the particle velocity $u$, and carries the momentum $F\tau$ of the force that made it. Its energy is shared equally between kinetic and elastic.
speeds of the source and the receiver relative to the medium
A moving source squeezes the wavelength ahead of it to $(c - v)/\nu_0$, a moving receiver meets crests more often. Signs are chosen so that approach raises the frequency. Reflection from a moving object shifts twice.
$\displaystyle \nu_n = \frac{nc}{2l}\ \text{(струна, труба с двумя открытыми концами)}$$\displaystyle \nu_n = \frac{(2n+1)c}{4l}\ \text{(труба, закрытая с одного конца)}$
A
amplitude of each of the two counter-running waves
l
length of the string or pipe
n
harmonic number
The incident and reflected waves add to a standing wave in which every point oscillates with a fixed amplitude, nodes and antinodes stay put, and neighbouring nodes are half a wavelength apart. A fixed end of a string and a closed end of a pipe give a node, an open end an antinode, so the length holds a whole number of half waves or an odd number of quarters.
particle velocities in the incident, reflected and transmitted waves
$\rho_1 c_1, \rho_2 c_2$
wave impedances of the media
At the boundary the particle velocity and the pressure are continuous, which gives the reflected and transmitted fractions through the wave impedances. A stiffer medium reflects with an inverted displacement, a softer one without, and equal $\rho c$ means no reflection. The frequency is kept across the boundary while the wavelength changes with the speed.
$$I = \frac{1}{2}\rho c\,\omega^2 A^2, \qquad I = \frac{P}{4\pi r^2}$$
$\displaystyle I = \frac{\Delta p_{\max}^2}{2\rho c}$$\displaystyle w = \frac{1}{2}\rho\omega^2 A^2\ \text{(плотность энергии)}$
I
intensity, energy per unit area per unit time
A
displacement amplitude of the particles
P
power of the source
r
distance from a point source
The energy density of a wave is $\rho\omega^2A^2/2$, and a volume $c$ crosses a unit area every second, hence the intensity. A point source spreads its power over a sphere, so the intensity falls as $1/r^2$ and the amplitude as $1/r$. Through the pressure amplitude the intensity is $\Delta p_{\max}^2/2\rho c$.