Oscillations and Waves

A tuning fork of frequency $f$ is carried towards a wall at speed $u$. An observer stands behind the fork, so the fork is moving away from them. They hear beats. Where do the beats come from?

There is only one fork, so there must be two frequencies reaching the ear. One is the sound travelling directly backwards from a receding source. The other has bounced off the wall.

The wall Dopplers twice. It first acts as a stationary observer receiving from an approaching source, and then re-radiates what it received as a stationary source. So

$f_{direct}=\frac{fv}{v+u},f_{reflected}=\frac{fv}{v-u}$

$f_{beat}=fv\left(\frac{1}{v-u}-\frac{1}{v+u}\right)=\frac{2fuv}{v^2-u^2}\approx \frac{2fu}{v}$

Nothing in that argument is a remembered special case. That is the difference this chapter is built on. Main gives four standard oscillators and one Doppler formula. Advanced gives a potential energy function and asks for the frequency, gives a junction and asks what fraction reflects, gives a reflector and expects you to apply the shift twice.

1. Any smooth minimum is a harmonic oscillator

Expand a potential about a minimum at $x_0$. The linear term vanishes there, so

$U(x)\approx U(x_0)+\frac{1}{2}{U}^{\prime \prime }(x_0){\left(x-x_0\right)}^2\Rightarrow \omega =\sqrt{\frac{U^{\prime \prime }(x_0)}{m}}$

This single line replaces every separate derivation. The rotational version is $\omega =\sqrt{C/I}$ for a restoring torque $\tau =-C\theta$.

Illustration 1

A particle of mass $m$ moves in the potential $U(x)=\frac{a}{x^2}-\frac{b}{x}$. Find the equilibrium position and the frequency of small oscillations about it.

$U^{\prime }=-\frac{2a}{x^3}+\frac{b}{x^2}=0\Rightarrow x_0=\frac{2a}{b}$

$U^{\prime \prime }=\frac{6a}{x^4}-\frac{2b}{x^3}$, and substituting $x_0$ gives $U^{\prime \prime }(x_0)=\frac{b^4}{8a^3}$

$\omega =\sqrt{\frac{b^4}{8a^{3}m}}$

Check the sign before trusting it. $U^{\prime \prime }>0$ confirms a genuine minimum; a negative value would mean an unstable equilibrium, where a displaced particle runs away instead of oscillating.

Illustration 2

A bead slides without friction on a wire bent to the shape $y=\frac{x^2}{4a}$. Find the period of small oscillations about the bottom.

$U=mgy=\frac{mg{x}^2}{4a}$, so $U^{\prime \prime }=\frac{mg}{2a}$

$\omega =\sqrt{\frac{g}{2a}}\Rightarrow T=2\pi \sqrt{\frac{2a}{g}}$

The wire behaves as a pendulum of length $2a$, which is exactly the radius of curvature of the parabola at its vertex. Any smooth valley oscillates like a pendulum of its local radius of curvature.

2. Physical and torsional pendulums

A rigid body swinging about a pivot a distance $d$ from its centre of mass has $I=I_{cm}+m{d}^2=m\left(k^2+d^2\right)$, so

$T=2\pi \sqrt{\frac{I}{mgd}}=2\pi \sqrt{\frac{k^2+d^2}{gd}},L_{eq}=\frac{k^2+d^2}{d}$

The equivalent length is what a simple pendulum of the same period would have. Minimising $T$ over $d$ gives the striking result

$d=k\Rightarrow T_{min}=2\pi \sqrt{\frac{2k}{g}}$

Because $L_{eq}$ is quadratic in $d$, two different pivot distances give the same period, with $d_1{d}_2=k^2$ and $d_1+d_2=L_{eq}$. That pair is what makes Kater's pendulum an accurate way to measure $g$.

Illustration 3

A uniform rod of length $L$ is pivoted at one end. Find its period, and the pivot position that minimises it.

$k^2=L^2/12$ and $d=L/2$:

$T=2\pi \sqrt{\frac{L^2/12+L^2/4}{gL/2}}=2\pi \sqrt{\frac{2L}{3g}}$

Minimum at $d=k=L/\sqrt{12}=0.289L$ from the centre, giving $T_{min}=2\pi \sqrt{0.577L/g}$.

Note that $T$ rises again as the pivot approaches the centre of mass, because the restoring torque vanishes there. A rod pivoted exactly at its centre does not oscillate at all.

Illustration 4

A body has the same period about two pivots at distances $0.3$ m and $0.7$ m from its centre of mass, the common period being $2.0$ s. Find $g$ and the radius of gyration.

$L_{eq}=d_1+d_2=1.0$ m and $k^2=d_1{d}_2=0.21$ m${}^2$, so $k=0.458$ m.

$g=\frac{4{\pi }^2{L}_{eq}}{T^2}=\frac{4{\pi }^2\times 1.0}{4.0}=9.87$ m s${}^{-2}$

The method's power is that $k$ never has to be measured. Only the distance between two pivots and one period are needed, which is why Kater's design dominated gravimetry for a century.

3. Two-body oscillation, and springs that are not ideal

When both ends of a spring are free to move, the separation obeys SHM with the reduced mass:

$\omega =\sqrt{\frac{k}{\mu }},\mu =\frac{m_1{m}_2}{m_1+m_2}$

The two blocks oscillate about a stationary centre of mass with amplitudes inversely proportional to their masses. A spring of its own mass $m_s$ contributes an effective inertia of $m_s/3$, since the coils near the fixed end barely move. And a spring cut into $n$ equal pieces has each piece $n$ times stiffer.

Illustration 5

Blocks of $2$ kg and $3$ kg are joined by a spring of stiffness $k=600$ N m${}^{-1}$ on frictionless ice. Find the frequency and the ratio of their amplitudes.

$\mu =\frac{2\times 3}{5}=1.2$ kg

$\omega =\sqrt{\frac{600}{1.2}}=22.4$ rad s${}^{-1}$, so $f=3.56$ Hz

$\frac{A_1}{A_2}=\frac{m_2}{m_1}=\frac{3}{2}$

The centre of mass never moves, because no external force acts. That is the fastest way to get the amplitude ratio and the reason the reduced mass appears at all.

Illustration 6

A spring of stiffness $k$ and mass $m_s=0.3$ kg carries a block of $1.2$ kg. Find the fractional error in the period if the spring's mass is ignored.

Effective mass $=1.2+\frac{0.3}{3}=1.3$ kg.

$\frac{T_{true}}{T_{ideal}}=\sqrt{\frac{1.3}{1.2}}=1.041$

The period is underestimated by about $4\%$. Because the correction sits under a square root, even a spring a quarter as heavy as the block shifts the period by only a few per cent.

4. Superposition of two simple harmonic motions

Along the same line, same frequency, the two motions add as phasors:

$A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

Along perpendicular directions, same frequency, the path is in general an ellipse. It degenerates into a straight line at $\varphi =0$ or $\pi$, and becomes a circle when $\varphi =\pi /2$ and the amplitudes are equal. Unequal frequencies produce Lissajous figures whose shape encodes the frequency ratio.

Illustration 7

A particle has $x=A\sin \omega t$ and $y=A\sin \left(\omega t+\pi /6\right)$. Identify the path.

Equal amplitudes with $\varphi =\pi /6$, which is neither $0$, $\pi$ nor $\pi /2$, so the path is a tilted ellipse inscribed in a square of side $2A$.

Eliminating $t$ gives $x^2+y^2-2xy\cos \frac{\pi }{6}=A^2{\sin }^2\frac{\pi }{6}$, an ellipse with its major axis along $y=x$.

Equal amplitudes alone do not give a circle. The phase difference must be exactly a quarter cycle as well, and any other value tilts and squashes the ellipse.

5. Damping, forcing and resonance

With a resistive force $-b\dot{x}$, the equation $m\ddot{x}+b\dot{x}+kx=0$ gives a decaying oscillation:

$x=A_0{e}^{-bt/2m}\cos {\omega }^{\prime }t,{\omega }^{\prime }=\sqrt{{\omega }_0^2-\frac{b^2}{4m^2}}$

Amplitude decays with time constant $2m/b$, and since energy goes as amplitude squared, energy decays twice as fast. The quality factor measures how many oscillations survive:

$Q=\frac{{\omega }_{0}m}{b}=2\pi \frac{\text{energy stored}}{\text{energy lost per cycle}}$

Under a driving force $F_0\cos \omega t$ the steady-state amplitude is

$A=\frac{F_0/m}{\sqrt{{\left({\omega }_0^2-{\omega }^2\right)}^2+{\left(b\omega /m\right)}^2}}$

which peaks near ${\omega }_0$. A high $Q$ makes that peak both taller and sharper.

Illustration 8

A damped oscillator's amplitude falls to half its initial value in $20$ oscillations. Find the fraction of energy remaining after $20$ oscillations, and estimate $Q$.

Energy $\propto A^2$, so the remaining fraction is ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$.

Per cycle the amplitude ratio is $2^{-1/20}$, so the energy ratio is $2^{-1/10}=0.933$, a loss of $6.7\%$.

$Q\approx \frac{2\pi }{0.067}=94$

A useful reflex: an amplitude that halves in $n$ cycles gives $Q\approx \frac{2\pi n}{2\ln 2}=\frac{\pi n}{\ln 2}$, which reproduces $91$ here.

6. Waves on a string: power, reflection and transmission

A travelling wave on a string carries

$P=\frac{1}{2}\mu {\omega }^2{A}^{2}v,v=\sqrt{\frac{T}{\mu }}$

so power depends on the square of both amplitude and frequency. At a junction between two strings, the amplitude reflection and transmission coefficients are

$r=\frac{v_2-v_1}{v_2+v_1},t=\frac{2v_2}{v_2+v_1}$

Going into a denser string means $v_2<v_1$, so $r$ is negative — the reflected pulse is inverted. Going into a rarer string leaves it upright. The fractions of power are $r^2$ reflected and $1-r^2$ transmitted.

Illustration 9

A wave travels from a string of linear density $\mu$ into one of density $4\mu$ under the same tension. Find the amplitude coefficients and the fractions of power reflected and transmitted.

$v_2=v_1/2$, so $r=\frac{0.5v_1-v_1}{1.5v_1}=-\frac{1}{3}$ and $t=\frac{2\left(0.5v_1\right)}{1.5v_1}=\frac{2}{3}$

Reflected power fraction $=r^2=\frac{1}{9}$; transmitted $=\frac{8}{9}$.

The transmitted amplitude is larger than the incident one is impossible to reconcile with energy — until you notice it is smaller here. Going the other way, into a rarer string, $t=4/3$ exceeds one, and the power still balances because the lighter string carries less energy per unit amplitude.

7. Particle velocity, intensity and interference

A snapshot of a wave and a movie of one molecule are different things. Differentiating $y=f\left(x-vt\right)$ shows that the transverse velocity of a particle is set by the slope of the string at that instant:

$v_{particle}=-v_{wave}\frac{\partial y}{\partial x}$

so particles at a crest or trough are momentarily at rest, and those crossing the axis move fastest. The particle speed is entirely unrelated in size to the wave speed.

For a source radiating uniformly in three dimensions, energy spreads over a sphere, so

$I=\frac{P}{4\pi r^2}\propto \frac{1}{r^2},A\propto \frac{1}{r},\beta =10{\log }_{10}\frac{I}{I_0}\text{dB}$

Two coherent sources produce a resultant intensity that depends on the phase difference rather than on a simple sum:

$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \delta ,\delta =\frac{2\pi }{\lambda }\Delta x$

Maxima occur where the path difference is a whole number of wavelengths, minima at odd half-wavelengths.

Illustration 10

A wave $y=0.02\sin \left(4x-200t\right)$ (SI units) travels on a string. Find the wave speed and the maximum particle speed, and state where along the wave each particle moves fastest.

$v_{wave}=\frac{\omega }{k}=\frac{200}{4}=50$ m s${}^{-1}$

$v_{particle,max}=\omega A=200\times 0.02=4$ m s${}^{-1}$

The two differ by a factor of more than twelve, and the particles move fastest exactly where the string crosses its equilibrium position, since that is where the slope is steepest.

Illustration 11

Two loudspeakers driven in phase are $3.0$ m apart. A listener stands $4.0$ m directly in front of one of them. Find the lowest frequency at which they cancel, taking $v=340$ m s${}^{-1}$.

Distances are $4.0$ m and $\sqrt{3^2+4^2}=5.0$ m, so $\Delta x=1.0$ m.

Cancellation needs $\Delta x=\lambda /2$, so $\lambda =2.0$ m.

$f=\frac{340}{2.0}=170$ Hz

Complete cancellation also requires equal amplitudes, which the unequal distances prevent. In practice the minimum is deep but not silent.

8. Standing waves, end correction and beats

An open pipe supports $f_n=\frac{nv}{2L}$; a closed pipe supports only odd harmonics, $f_n=\frac{(2n-1)v}{4L}$. Real pipes have an end correction of about $0.6r$ at each open end, because the antinode sits slightly outside.

The resonance-tube experiment removes the correction by taking two resonances:

$v=2f\left(l_2-l_1\right),e=\frac{l_2-3l_1}{2}$

Two nearby frequencies give beats at $\left|f_1-f_2\right|$. Loading a fork with wax lowers its frequency, and observing whether the beat rate rises or falls is what identifies which fork was which.

Illustration 12

A resonance tube with a fork of $480$ Hz resonates at column lengths $16.0$ cm and $50.0$ cm. Find the speed of sound and the end correction.

$v=2f\left(l_2-l_1\right)=2(480)(0.34)=326$ m s${}^{-1}$

$e=\frac{l_2-3l_1}{2}=\frac{50.0-48.0}{2}=1.0$ cm

Using only the first resonance would give $v=4f{l}_1=307$ m s${}^{-1}$, low by $6\%$. The two-resonance method exists precisely because the end correction is not negligible.

Illustration 13

Two forks give $5$ beats per second. Waxing the first fork raises the beat rate to $8$ per second. Which fork had the higher frequency, and what was it, given the second is $256$ Hz?

Waxing lowers a fork's frequency. The gap widened, so the waxed fork was already the lower one and moved further away.

The first fork is therefore $256-5=251$ Hz, and the second, at $256$ Hz, is the higher.

The direction of the change is the entire question. Beat frequency alone gives two candidates, and only a deliberate perturbation resolves which is which.

9. Doppler done properly

For motion along the line joining source and observer,

$f^{\prime }=f\frac{v+v_o}{v-v_s}$

with $v_o$ positive when the observer moves towards the source and $v_s$ positive when the source moves towards the observer. Three refinements carry the Advanced marks.

Only the component along the line counts. A source passing at closest approach has zero radial velocity, so the shift is momentarily zero even though the speed is largest.

Wind adds to $v$ in the direction of propagation, appearing in both numerator and denominator — so a wind blowing from source to observer does not change the observed frequency for stationary source and observer.

A reflector Dopplers twice, once as observer and once as source.

Illustration 14

A fork of $340$ Hz moves towards a wall at $2$ m s${}^{-1}$. An observer behind the fork hears beats. Find the beat frequency, with $v=340$ m s${}^{-1}$.

$f_{direct}=\frac{340\times 340}{342}=338.0$ Hz

$f_{reflected}=\frac{340\times 340}{338}=342.0$ Hz

$f_{beat}=4.0$ Hz, matching the estimate $\frac{2fu}{v}=\frac{2\times 340\times 2}{340}=4$ Hz.

The approximation is excellent because $u\ll v$. At everyday speeds the exact and approximate forms agree to well within the precision of any measurement you could make by ear.

Illustration 15

A source moves in a circle at constant speed while an observer stands outside the circle. When is the observed frequency highest, lowest, and equal to the true frequency?

Highest when the source moves directly towards the observer, which happens at a point where the tangent to the circle points at them.

Lowest at the diametrically related point where the tangent points directly away.

Equal to $f$ at the two points of closest and farthest approach, where the velocity is entirely transverse and the radial component is zero.

The naive answer places the extremes at closest approach. It is the radial component, not the distance, that produces the shift.

Summary

• Any smooth potential minimum is harmonic for small displacements: $\omega =\sqrt{U^{\prime \prime }(x_0)/m}$, and $\omega =\sqrt{C/I}$ for torques.
• Physical pendulum: $T=2\pi \sqrt{\frac{k^2+d^2}{gd}}$, equivalent length $L_{eq}=\frac{k^2+d^2}{d}$.
• $T$ is minimum at $d=k$, and two pivot distances share a period with $d_1{d}_2=k^2$, $d_1+d_2=L_{eq}$ — the basis of Kater's pendulum.
• Two free masses on a spring: $\omega =\sqrt{k/\mu }$ with $\mu =\frac{m_1{m}_2}{m_1+m_2}$, amplitudes inversely as the masses.
• A spring's own mass contributes $m_s/3$; cutting a spring into $n$ pieces makes each $n$ times stiffer.
• Collinear superposition adds as phasors: $A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$.
• Perpendicular superposition gives an ellipse; a line at $\varphi =0,\pi$ and a circle only at $\varphi =\pi /2$ with equal amplitudes.
• Damping: amplitude decays as $e^{-bt/2m}$, energy twice as fast; $Q={\omega }_{0}m/b$ counts surviving oscillations.
• Wave power $P=\frac{1}{2}\mu {\omega }^2{A}^{2}v$ — quadratic in both amplitude and frequency.
• Junction: $r=\frac{v_2-v_1}{v_2+v_1}$, $t=\frac{2v_2}{v_2+v_1}$; entering a denser string inverts the reflected pulse.
• Power fractions are $r^2$ and $1-r^2$; a transmitted amplitude above one is allowed, since the lighter string carries less energy per unit amplitude.
• $v_{particle}=-v_{wave}\partial y/\partial x$: particles are fastest at the axis crossings, slowest at the crests.
• Spherical spreading gives $I\propto 1/r^2$ and $A\propto 1/r$; two coherent sources give $I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \delta$.
• Resonance tube: $v=2f\left(l_2-l_1\right)$ and $e=\frac{l_2-3l_1}{2}$ — two resonances remove the end correction.
• Doppler: only the radial component shifts the frequency; wind cancels between numerator and denominator; a reflector shifts twice, giving $f_{beat}\approx 2fu/v$.