2.P — Worked Problems: Oscillations, Waves and Sound

Twelve problems across Part 2, climbing from standard to genuinely awkward. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — Reading SHM from its equation

A particle moves according to x = 0.15\sin(8\pi t + \pi/3) metres, with t in seconds. Find the amplitude, frequency, period, maximum speed, maximum acceleration, and the position at t = 0.

Solution

Compare with x = A\sin(\omega t+\phi).

A = 0.15\ \text{m}, \qquad \omega = 8\pi = 25.13\ \text{rad/s}, \qquad \phi = \pi/3

f = \frac{\omega}{2\pi} = \frac{8\pi}{2\pi} = 4.0\ \text{Hz}, \qquad T = \frac{1}{4.0} = 0.25\ \text{s}

v_{\max} = A\omega = 0.15\times25.13 = 3.77\ \text{m/s}

a_{\max} = A\omega^2 = 0.15\times631.7 = 94.7\ \text{m/s}^2

At t=0:

x = 0.15\sin(\pi/3) = 0.15\times0.866 = 0.13\ \text{m}

What to notice. The maximum acceleration is 9.7 times g, from an oscillation only 15 cm in amplitude. Acceleration carries \omega^2, so anything oscillating quickly is subject to violent accelerations however small its motion — which is why a loudspeaker cone at 4 kHz survives hundreds of g and why fatigue, not force, is what kills vibrating machinery.

Problem 2 — Two springs, two ways

A 2.0 kg mass is connected to two springs of stiffness 300 N/m and 600 N/m. Find the period when they are (a) in parallel, both pulling on the mass, and (b) in series, one after the other.

Solution

(a) Parallel. Both springs stretch by the same x, and their forces add:

F = k_1x+k_2x = (k_1+k_2)x \quad\Longrightarrow\quad k_{\text{eff}} = 300+600 = 900\ \text{N/m}

T = 2\pi\sqrt{\frac{2.0}{900}} = 2\pi\sqrt{0.002222} = 2\pi\times0.04714 = 0.296\ \text{s}

(b) Series. Both springs carry the same force F, and the extensions add:

x = x_1+x_2 = \frac{F}{k_1}+\frac{F}{k_2} = F\left(\frac{1}{k_1}+\frac{1}{k_2}\right)

\frac{1}{k_{\text{eff}}} = \frac{1}{300}+\frac{1}{600} = \frac{2}{600}+\frac{1}{600} = \frac{3}{600} \quad\Longrightarrow\quad k_{\text{eff}} = 200\ \text{N/m}

T = 2\pi\sqrt{\frac{2.0}{200}} = 2\pi\times0.1 = 0.628\ \text{s}

What to notice. The rules are the reverse of resistors: springs in parallel add directly, springs in series add reciprocally. The reason is which quantity is shared — parallel springs share the displacement, series springs share the force. Note also that the series pair is softer than either spring alone, which is why a long spring is easier to stretch than a short one cut from it: a long spring is many short ones in series.

Problem 3 — Damping from observation

A pendulum's amplitude falls from 12 cm to 3.0 cm in 40 s. Find the damping rate \gamma, the Q factor if its period is 2.0 s, and how long until the amplitude is 1% of its start.

Solution

The envelope is A(t) = A_0e^{-\gamma t}:

\frac{3.0}{12} = e^{-40\gamma} \quad\Longrightarrow\quad 0.25 = e^{-40\gamma}

Take natural logs:

\ln 0.25 = -40\gamma \quad\Longrightarrow\quad -1.386 = -40\gamma \quad\Longrightarrow\quad \gamma = 0.03466\ \text{s}^{-1}

\omega_0 = \frac{2\pi}{T} = \frac{6.283}{2.0} = 3.142\ \text{rad/s}

Q = \frac{\omega_0}{2\gamma} = \frac{3.142}{0.06931} = 45.3

For 1% of the start:

0.01 = e^{-\gamma t} \quad\Longrightarrow\quad t = \frac{\ln 100}{\gamma} = \frac{4.605}{0.03466} = 133\ \text{s}

What to notice. Q \approx 45 means the pendulum makes roughly 45 oscillations before dying to about 4% — and 133 s at 2 s per swing is 66 swings to reach 1%, which is consistent. Also worth checking: is the frequency shift from damping significant? \omega_d = \sqrt{\omega_0^2-\gamma^2} = \sqrt{9.87-0.0012} = 3.1416, unchanged to five figures. With Q in the tens, the damped and undamped frequencies are indistinguishable, which is exactly why a dying pendulum still keeps time.

Problem 4 — Resonance and a car's suspension

A 1200 kg car sits on four springs, and its body drops 8.0 cm when the car is loaded. Find the natural frequency. Then find the road speed at which the car resonates on a road with bumps every 6.0 m.

Solution

Effective stiffness from the static droop. At equilibrium the springs support the weight:

k_{\text{total}} = \frac{mg}{x} = \frac{1200\times9.81}{0.080} = \frac{11\,772}{0.080} = 147\,150\ \text{N/m}

\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{147\,150}{1200}} = \sqrt{122.6} = 11.07\ \text{rad/s}

f_0 = \frac{11.07}{2\pi} = 1.76\ \text{Hz}

Resonant speed. Bumps arrive at frequency v/L:

\frac{v}{6.0} = 1.76 \quad\Longrightarrow\quad v = 10.6\ \text{m/s} = 38\ \text{km/h}

What to notice. There is a shortcut hiding in the first step. Substituting k = mg/x into \omega = \sqrt{k/m} gives \omega = \sqrt{g/x} — the mass cancels, and the natural frequency of any spring system depends only on how far it sags under its own weight. That is the pendulum formula in disguise, with the static droop playing the role of the length, and it lets you estimate a suspension's frequency by pushing down on the wing and watching.

A resonance at 38 km/h on a road with regular joints is exactly the situation that produces the alarming pitching some cars do at one particular speed, and it is why shock absorbers are set near critical damping (Chapter 2.2).

Problem 5 — Guitar string, from tension to pitch

A steel guitar string is 0.65 m long between the nut and bridge, has a mass of 3.2 g per metre, and is tuned to 110 Hz (low A). Find the tension. Then find the fret position for 146.8 Hz (D).

Solution

The fundamental of a fixed-fixed string is f_1 = \frac{1}{2L}\sqrt{F/\mu}. Rearranging:

F = \mu(2Lf_1)^2 = 0.0032\times(2\times0.65\times110)^2

2Lf_1 = 143.0\ \text{m/s} \quad\text{(this is the wave speed on the string)}

F = 0.0032\times143.0^2 = 0.0032\times20\,449 = 65.4\ \text{N}

About 6.7 kg of pull, from one string. A six-string guitar therefore pulls its neck with roughly 400 N, which is why necks have a steel truss rod inside them.

Fret position. Tension and \mu are unchanged, so v = 143.0 m/s stays fixed and only L changes:

L = \frac{v}{2f} = \frac{143.0}{2\times146.8} = \frac{143.0}{293.6} = 0.487\ \text{m}

So the fret sits 0.65-0.487 = 0.163 m from the nut.

Check against the equal-tempered rule. Each fret divides the length by 2^{1/12} = 1.0595. D is five semitones above A, so:

L = \frac{0.65}{1.0595^5} = \frac{0.65}{1.3348} = 0.487\ \text{m}$$ ✓ **What to notice.** The frets get closer together as you go up the neck, because the length is divided by a constant *ratio* each time, not reduced by a constant amount. That geometric spacing is the physical signature of equal temperament, and you can see it on any guitar.

Problem 6 — The pipe that is closed at one end

An organ pipe closed at one end is 1.20 m long. Find its first three resonant frequencies at 20 °C. Then find them at 5 °C.

Solution

At 20 °C, v = 343 m/s. A closed pipe gives odd harmonics only, with f_n = nv/4L:

f_1 = \frac{343}{4\times1.20} = \frac{343}{4.80} = 71.5\ \text{Hz}

f_3 = 3\times71.5 = 214.5\ \text{Hz}, \qquad f_5 = 5\times71.5 = 357.5\ \text{Hz}

At 5 °C, v = 331+0.6\times5 = 334\ \text{m/s}:

f_1 = \frac{334}{4.80} = 69.6\ \text{Hz}

How flat is that? The ratio is 69.6/71.5 = 0.9734. In semitones:

n = 12\log_2(0.9734) = 12\times\frac{\ln 0.9734}{\ln 2} = 12\times\frac{-0.02695}{0.6931} = -0.47\ \text{semitones}

Nearly half a semitone flat. Painfully audible.

What to notice. The pipe's length did not change; only the air did. This is why a pipe organ must be tuned at its playing temperature and why a cold church makes the organ sound wrong even when nothing is broken. Note also that stringed instruments go the other way in the cold — the string contracts and the tension rises, sharpening them — which is why an orchestra in a cold hall goes progressively out of tune with the organ in two directions at once.

Problem 7 — Beats, and finding which string is wrong

A tuning fork of 440 Hz is sounded with a string. Four beats per second are heard. The string is then tightened slightly and the beat rate rises to seven per second. What was the string's original frequency?

Solution

Four beats means the string was at 440\pm4, so either 436 or 444 Hz. One more piece of information is needed to decide, and tightening supplied it.

Tightening raises the string's frequency, since f\propto\sqrt{F}.

• If the string was at 444, tightening moves it further from 440 and the beat rate should rise. ✓
• If the string was at 436, tightening moves it towards 440 and the beat rate should fall. ✗

So the string was at 444 Hz, and after tightening it is at 447 Hz.

What to notice. The beat rate alone is an absolute value and can never tell you the sign. You always need a second measurement that changes one of the frequencies in a known direction. Every piano tuner uses this: they deliberately push the string slightly and watch whether the beating quickens or slows. The same ambiguity and the same fix appear in radio, radar and interferometry.

Problem 8 — Two speakers and a quiet line

Two speakers 2.0 m apart emit the same 1.2 kHz tone in phase. You walk along a line 5.0 m in front of them, parallel to the line joining them. How far from the centre is the first quiet spot?

Solution

\lambda = \frac{343}{1200} = 0.2858\ \text{m}

The first minimum needs a path difference of half a wavelength, \Delta L = 0.1429 m.

Let y be the sideways distance from the centre line. The two path lengths are:

L_1 = \sqrt{5.0^2+(y-1.0)^2}, \qquad L_2 = \sqrt{5.0^2+(y+1.0)^2}

For a distant screen the standard approximation is \Delta L \approx dy/D with d = 2.0 m and D = 5.0 m:

\frac{2.0\,y}{5.0} = 0.1429 \quad\Longrightarrow\quad y = \frac{0.1429\times5.0}{2.0} = 0.357\ \text{m}

Check the approximation is fair. Compute the exact paths at y = 0.357:

L_1 = \sqrt{25+(0.357-1.0)^2} = \sqrt{25+0.4134} = \sqrt{25.413} = 5.0412

L_2 = \sqrt{25+(1.357)^2} = \sqrt{25+1.8414} = \sqrt{26.841} = 5.1808

\Delta L = 0.1396\ \text{m}

against the required 0.1429 — about 2% out, so the approximation is good and the exact answer is very slightly further out, around 0.366 m.

What to notice. The quiet spots at this frequency are roughly 70 cm apart, so moving your head by 35 cm takes you from loud to quiet. At 120 Hz instead, \lambda = 2.9 m and the spacing is ten times larger, which is why bass fills a room evenly and treble is fussy about where you sit.

Problem 9 — Doppler, both moving, with a wind

A train sounds a 400 Hz horn while moving at 30 m/s towards a stationary observer. A wind blows at 10 m/s from the train towards the observer. What frequency is heard? Take still-air sound speed as 340 m/s.

Solution

Wind does not change the frequency by itself — but it changes the effective speed of sound relative to the ground, which is the frame both the formula's speeds are measured in.

With the wind blowing from source to observer, sound travels over the ground at:

v_{\text{eff}} = 340+10 = 350\ \text{m/s}

Now apply the source formula with this effective speed:

f_o = f_s\frac{v_{\text{eff}}}{v_{\text{eff}}-v_s} = 400\times\frac{350}{350-30} = 400\times\frac{350}{320} = 437.5\ \text{Hz}

Compare with no wind:

f_o = 400\times\frac{340}{310} = 438.7\ \text{Hz}

What to notice. The wind changed the answer by only 1.2 Hz, and in the direction most people guess wrong — a tailwind slightly lowers the observed shift. The reason is that raising v raises the numerator and the denominator together, and since the denominator is the smaller number, the proportional effect there is larger. A wind changes what you hear far less than it changes how far you hear it, because its main effect is the refraction described in Chapter 2.3, not the Doppler shift.

Problem 10 — The supersonic aircraft overhead

An aircraft flies level at 12 km altitude at Mach 1.8. Sound speed at that altitude is 295 m/s. How far past you is it when you hear the boom?

Solution

\sin\theta = \frac{1}{M} = \frac{1}{1.8} = 0.5556 \quad\Longrightarrow\quad \theta = 33.75°

The shock cone trails behind the aircraft with half-angle \theta measured from its flight path. You hear the boom when the cone's surface reaches you. Geometrically, if the aircraft is at altitude h, the cone meets the ground at a horizontal distance d behind it where:

\tan\theta = \frac{h}{d} \quad\Longrightarrow\quad d = \frac{h}{\tan\theta} = \frac{12\,000}{\tan 33.75°} = \frac{12\,000}{0.6682} = 17\,960\ \text{m}

About 18 km.

How long after it passed overhead? The aircraft's ground speed is 1.8\times295 = 531 m/s:

t = \frac{17\,960}{531} = 33.8\ \text{s}

What to notice. You hear the boom 34 seconds after the aircraft has gone by, when it is 18 km away and probably out of sight. This is why witnesses consistently report the boom as coming from empty sky, and why "the boom happens when it breaks the sound barrier" is wrong — the cone trails continuously, and it sweeps across the ground as a moving carpet roughly 40 km wide for an aircraft at this height.

Problem 11 — The tricky one: a mass between two walls

A 0.40 kg block sits on a frictionless surface, connected by identical springs (k = 250 N/m each) to two walls. Both springs are at natural length when the block is centred. The block is displaced 5.0 cm and released. Find the period.

Then repeat for the case where both springs are already stretched by 10 cm when the block is centred.

Solution

Case 1: springs at natural length. Displace the block right by x. The right spring is compressed by x and pushes left with kx; the left spring is stretched by x and pulls left with kx. Both act in the same direction:

F = -2kx \quad\Longrightarrow\quad k_{\text{eff}} = 500\ \text{N/m}

T = 2\pi\sqrt{\frac{0.40}{500}} = 2\pi\sqrt{0.0008} = 2\pi\times0.02828 = 0.178\ \text{s}

Case 2: both springs pre-stretched by 10 cm. The instinct is that pre-tension must change something. Work it out.

At the centre, each spring is stretched 0.10 m and pulls with 25 N, in opposite directions, cancelling. Displace right by x:

• Right spring stretch becomes 0.10-x, pulling right with k(0.10-x).
• Left spring stretch becomes 0.10+x, pulling left with k(0.10+x).

Net force, taking right as positive:

F = k(0.10-x) - k(0.10+x) = 0.10k - kx - 0.10k - kx = -2kx

Identical. The pre-tension cancels exactly, and the period is 0.178 s again.

What to notice. This is the same result as the vertical spring in Chapter 2.1: a constant force added to a harmonic system moves the equilibrium and leaves the frequency alone. Here the two pre-tensions are equal and opposite so the equilibrium does not even move.

The catch, and it is the real point of the problem. Case 2 only behaves this way while both springs remain stretched. Once x > 0.10 m the right spring goes slack, and a real spring cannot push — it can only pull. Past that point only the left spring acts, F = -kx - constant, and the motion is no longer symmetric SHM. With an amplitude of 5.0 cm we are safely inside the linear region; at 15 cm the problem becomes a different one entirely, and the period would be longer. Always check that the assumed regime survives the amplitude you were given.

Problem 12 — Standing waves on a string with a hanging mass

A string of linear density 1.5 g/m runs horizontally from a vibrator over a pulley, with a mass M hanging from the free end. The horizontal section is 1.2 m long. The vibrator runs at 60 Hz. Find the masses that produce standing waves with one, two, and three loops.

Solution

The tension is the hanging weight, F = Mg. Both ends of the horizontal section are effectively fixed (the vibrator's amplitude is tiny, and the pulley is a fixed point), so:

f_n = \frac{n}{2L}\sqrt{\frac{F}{\mu}} \quad\Longrightarrow\quad \sqrt{\frac{Mg}{\mu}} = \frac{2Lf}{n}

Mg = \mu\left(\frac{2Lf}{n}\right)^2 \quad\Longrightarrow\quad M = \frac{\mu}{g}\left(\frac{2Lf}{n}\right)^2

With \mu = 0.0015 kg/m, L = 1.2 m, f = 60 Hz:

2Lf = 2\times1.2\times60 = 144\ \text{m/s}

One loop (n=1):

M = \frac{0.0015}{9.81}\times144^2 = 1.529\times10^{-4}\times20\,736 = 3.17\ \text{kg}

Two loops (n=2): the bracket becomes 144/2 = 72:

M = 1.529\times10^{-4}\times5184 = 0.793\ \text{kg}

Three loops (n=3): the bracket becomes 48:

M = 1.529\times10^{-4}\times2304 = 0.352\ \text{kg}

What to notice. The required masses fall as 1/n^2 — 3.17, 0.79, 0.35 kg — because tension enters the frequency as a square root, so hitting a mode n times higher needs n^2 times less tension. That is why the low-numbered modes are far apart in mass and the high ones crowd together, and why this experiment is easy to do for one and two loops and fiddly for five.

Note also what stayed fixed. The vibrator's 60 Hz never changed; the string's speed changed, and with it the wavelength. This is the v = f\lambda bookkeeping from Chapter 2.3 with the source holding f and the medium supplying v.