C.4 Standing waves and resonance – IB Physics exam questions with mark scheme

1Standard5 marks

A pipe of length $0.60\ \mathrm{m}$ is closed at one end. The speed of sound in air is $340\ \mathrm{m\,s^{-1}}$.

(a) Calculate the frequency of the fundamental (first harmonic) standing wave in the pipe. [2]

(b) Calculate the frequency of the next harmonic that can be set up. [1]

(d) State the nature of the displacement at the closed end of the pipe. [1]

Fundamental has a quarter wavelength in the pipe: $\lambda=4L=2.4$ m(1)
$f=v/\lambda = 340/2.4 = 142$ Hz (accept 140 Hz)(1)
Next harmonic is the third: $3f_1 = 425$ Hz (accept 430 Hz)(1)
$f=v/2L = 340/1.2 = 283$ Hz (accept 280 Hz)(1)
A (displacement) node at the closed end (accept “zero displacement”; reject “pressure node”)(1)
Answer: (a) 142 Hz (b) 425 Hz (c) 283 Hz (d) node
Examiner tip: A closed pipe supports only odd harmonics; the second harmonic does not exist.

2Core4 marks

A travelling wave and a standing wave can both be set up on a stretched rope.

(a) State two differences between a standing wave and a travelling wave. [2]

(b) Define a node of a standing wave. [1]

(c) State the phase difference between two points on a standing wave that lie on opposite sides of a single node. [1]

Any one difference, e.g. a standing wave does not transfer energy (along the rope) whereas a travelling wave does(1)
A second, different difference, e.g. in a standing wave the amplitude varies with position (zero at nodes, maximum at antinodes) whereas in a travelling wave all points have the same amplitude; OR points between adjacent nodes oscillate in phase whereas phase varies continuously along a travelling wave; OR the wave profile does not move along the rope. Do not credit two halves of the same idea twice(1)
A point on a standing wave where the displacement is (always) zero / amplitude is zero (accept point of no oscillation; ignore “no movement of energy”)(1)
$\pi$ rad / $180^\circ$ / antiphase (accept “half a cycle”)(1)
Answer: (a) e.g. no net energy transfer; amplitude varies with position (b) point of permanent zero displacement (c) $\pi$ rad
Examiner tip: Saying a standing wave “does not move” is not enough: say that its profile does not travel and that it transfers no energy.

3Standard6 marks

A string of length $1.20\ \mathrm{m}$ is fixed at both ends. It vibrates in the standing-wave pattern shown at a frequency of $150\ \mathrm{Hz}$. The solid line shows the string at one instant; the dashed line shows it half a period later. P, Q and R are points on the string.

(a) State the number of nodes on the string, including the ends. [1]

(b) Calculate the speed of the travelling waves on the string. [2]

(d) Calculate the time interval between two successive instants at which the whole string is straight. [1]

4(1)
$\lambda = \dfrac{2\times1.20}{3} = 0.80\ \mathrm{m}$ (allow 3 half-wavelengths fill 1.20 m)(1)
$v = f\lambda = 150\times0.80 = 120\ \mathrm{m\,s^{-1}}$ (ECF for wrong $\lambda$)(1)
$\pi$ rad / $180^\circ$ / antiphase(1)
zero / 0 rad / in phase (accept $2\pi$)(1)
$T/2 = \dfrac{1}{2\times150} = 3.3\times10^{-3}\ \mathrm{s}$ (accept 3.33 ms; do not accept $1/150$ s)(1)
Answer: (a) 4 (b) $120\ \mathrm{m\,s^{-1}}$ (c)(i) $\pi$ rad (ii) 0 (d) $3.3\ \mathrm{ms}$
Examiner tip: The string is straight twice per cycle, so the answer to (d) is half a period, not a whole one.

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4Core4 marks

The G string of a guitar has a vibrating length of $0.65\ \mathrm{m}$ and its fundamental (first harmonic) frequency is $196\ \mathrm{Hz}$.

(a) Calculate the speed of the transverse waves on the string. [2]

(b) A guitarist presses the string against a fret so that only part of it vibrates. The tension, and therefore the wave speed, is unchanged. Calculate the vibrating length needed to give a fundamental frequency of $247\ \mathrm{Hz}$. [2]

$\lambda = 2L = 1.30\ \mathrm{m}$(1)
$v = f\lambda = 196\times1.30 = 255\ \mathrm{m\,s^{-1}}$ (accept 254.8; allow 250)(1)
$L = \dfrac{v}{2f}$ or $L \propto \dfrac{1}{f}$ seen(1)
$L = \dfrac{255}{2\times247} = 0.52\ \mathrm{m}$ (accept 0.516 m; ECF from (a))(1)
Answer: (a) $255\ \mathrm{m\,s^{-1}}$ (b) $0.52\ \mathrm{m}$
Examiner tip: For a string fixed at both ends the fundamental wavelength is twice the length, not equal to it.

5Standard5 marks

A student investigates resonance in an air column. A vertical glass tube is lowered into a tall cylinder of water so that the length of air above the water can be changed. A tuning fork of frequency $512\ \mathrm{Hz}$ is struck and held above the open top of the tube.

Starting with a very short air column, the student slowly raises the tube. The first loud sound (resonance) is heard when the air column is $16.4\ \mathrm{cm}$ long and the second when it is $49.8\ \mathrm{cm}$ long.

(a) Explain why the difference between the two lengths is half a wavelength. [2]

(b) Determine the speed of sound in the air in the tube. [2]

(c) The antinode at the open end is actually slightly above the top of the tube. Use the data to estimate how far above. [1]

The water surface is a (displacement) node and the open end is (near) an antinode, so resonance occurs at lengths of about $\lambda/4$ and $3\lambda/4$(1)
Adding half a wavelength moves to the next node position / adds one more loop, so $L_2 - L_1 = \lambda/2$ (allow argument that end correction cancels in the difference)(1)
$\lambda = 2(0.498-0.164) = 0.668\ \mathrm{m}$(1)
$v = f\lambda = 512\times0.668 = 342\ \mathrm{m\,s^{-1}}$ (ECF for wrong $\lambda$)(1)
$\dfrac{\lambda}{4} - L_1 = 0.167 - 0.164 = 0.003\ \mathrm{m}$ (accept 0.3 cm; ECF from (b))(1)
Answer: (b) $342\ \mathrm{m\,s^{-1}}$ (c) about $0.3\ \mathrm{cm}$
Examiner tip: Using the difference of the two lengths removes the end correction; using $\lambda = 4L_1$ alone gives a less accurate speed.

6Standard4 marks

A student measures three consecutive resonant frequencies of an air-filled pipe: $150\ \mathrm{Hz}$, $250\ \mathrm{Hz}$ and $350\ \mathrm{Hz}$. The speed of sound in the air is $340\ \mathrm{m\,s^{-1}}$.

(a) Deduce whether the pipe is open at both ends or closed at one end. [2]

(b) Calculate the length of the pipe. [2]

Spacing between consecutive harmonics is $100\ \mathrm{Hz}$, so the fundamental is $50\ \mathrm{Hz}$ (allow: frequencies are not whole-number multiples of 100 Hz)(1)
The frequencies are 3, 5 and 7 times $50\ \mathrm{Hz}$ / only odd harmonics present, so the pipe is closed at one end (no mark for “closed” without reasoning)(1)
$L = \dfrac{\lambda}{4}$ with $\lambda = \dfrac{340}{50} = 6.8\ \mathrm{m}$ (or $L = \dfrac{v}{4f_1}$)(1)
$L = 1.7\ \mathrm{m}$ (ECF from wrong fundamental with consistent pipe type)(1)
Answer: (a) closed at one end (odd harmonics of 50 Hz) (b) $1.7\ \mathrm{m}$
Examiner tip: The lowest measured frequency is not necessarily the fundamental: use the spacing between harmonics first.

7Hard6 marks

The graph shows how the amplitude of a mass–spring system varies with the frequency of a periodic driving force. Curve A is for the system with light damping. Curve B is for the same system after the mass is placed in a viscous liquid.

(a) Estimate the natural frequency of the system. [1]

(b) Describe two ways in which curve B differs from curve A. [2]

(c) The driving force is removed while the system oscillates. Outline how the subsequent motion differs if the system is (i) under-damped and (ii) critically damped. [2]

(d) State why the suspension of a car is designed to be close to critically damped. [1]

$40\ \mathrm{Hz}$ (accept 38–42 Hz)(1)
Any one of: peak amplitude (much) smaller; resonance peak broader / flatter (less sharp); peak occurs at a (slightly) lower frequency(1)
A second, different one of these differences; do not credit “lower peak” and “smaller amplitude” (the same difference) as two separate points; ignore statements that the curves are similar away from resonance(1)
(i) Under-damped: system oscillates (about equilibrium) with amplitude decreasing (exponentially) over time(1)
(ii) Critically damped: returns to equilibrium in the shortest possible time without oscillating / overshooting(1)
So that after a bump the car returns to equilibrium quickly without bouncing / oscillating (reject “to stop resonance” alone)(1)
Answer: (a) 40 Hz (b) any two: lower peak; broader peak; peak at slightly lower frequency (c) decaying oscillations vs fastest return with no oscillation (d) to stop bouncing and settle quickly
Examiner tip: Heavy damping lowers and broadens the resonance peak but does not remove the natural frequency: the peak still occurs near $f_0$.

8Hard6 marks

A wire is stretched between two fixed clamps. A vibration generator near one end drives the wire at a frequency that can be varied.

Explain how a standing wave is formed on the wire, and why a large-amplitude standing wave is seen only at certain driving frequencies. Your answer should refer to nodes, antinodes and the phase of points on the wire. [6]

Waves travel along the wire and are reflected at the fixed ends(1)
The incident and reflected waves have the same frequency (and speed) and similar amplitude, travelling in opposite directions(1)
They superpose (interfere) to give a standing wave(1)
Nodes: points where the two waves are always in antiphase / cancel, giving zero displacement; antinodes where they are always in phase, giving maximum amplitude(1)
Fixed ends cannot move, so there must be a node at each end(1)
So a whole number of half-wavelengths must fit on the wire: $L = n\lambda/2$, giving $f_n = nv/(2L)$; only these frequencies are possible(1)
At these frequencies the driving frequency equals a natural frequency of the wire, so resonance occurs and the amplitude builds up (energy transfer from driver is maximum)(1)
Points between adjacent nodes oscillate in phase (with different amplitudes); points in adjacent loops are in antiphase(1)
Adjacent nodes (or antinodes) are $\lambda/2$ apart(1)

Level	Marks	What it looks like
Level 3	5–6	Coherent explanation linking reflection and superposition to node/antinode formation, with nodes at the fixed ends giving $L=n\lambda/2$, discrete frequencies explained as resonance, and correct phase relationships stated. Accurate terminology throughout.
Level 2	3–4	Describes formation by superposition of reflected waves and identifies nodes/antinodes; partial account of why only certain frequencies work (e.g. states harmonics without linking to fixed ends or resonance) or phase relationships missing or partly wrong.
Level 1	1–2	Isolated relevant points, e.g. waves reflect, or nodes do not move; little or no explanation of discrete frequencies; terminology loose.
Level 0	0	No relevant physics.

Answer: Indicative content: reflection and superposition of opposing waves; nodes at fixed ends; $L=n\lambda/2$; resonance at natural frequencies; in-phase within a loop, antiphase across a node.
Examiner tip: Many answers describe the pattern but never say why only certain frequencies work: link the fixed-end nodes to $L = n\lambda/2$ and to resonance.

9Stretch6 marks

The diagram shows the displacement pattern of a standing sound wave in a pipe of length $0.85\ \mathrm{m}$ that is open at both ends. The speed of sound in the air is $340\ \mathrm{m\,s^{-1}}$.

(a) State which harmonic is shown. [1]

(b) Calculate the frequency of this standing wave. [2]

(c) One end of the pipe is now closed. Show that none of the resonant frequencies of the closed pipe is equal to a resonant frequency of the original open pipe. [3]

Second harmonic (accept $n = 2$)(1)
$\lambda = L = 0.85\ \mathrm{m}$ (one full wavelength in the pipe)(1)
$f = \dfrac{340}{0.85} = 400\ \mathrm{Hz}$(1)
Open pipe: $f_n = \dfrac{nv}{2L} = 200n$ Hz, i.e. 200, 400, 600 … Hz (all even multiples of 100 Hz)(1)
Closed pipe: $f = \dfrac{(2m-1)v}{4L} = 100(2m-1)$ Hz, i.e. 100, 300, 500 … Hz (odd multiples of 100 Hz)(1)
An odd multiple of 100 Hz can never equal an even multiple, so no frequency is common (conclusion must be stated)(1)
Answer: (a) second (b) 400 Hz (c) open: 200n Hz; closed: 100(2m−1) Hz; even vs odd multiples of 100 Hz
Examiner tip: In (c) listing a few values is not a proof: write the general expressions and compare them.

10Stretch6 marks

A microwave transmitter faces a flat metal plate. A small detector is moved along the line between them and its reading rises and falls as it passes through a standing wave.

The detector is placed at a minimum. It is then moved towards the plate until it has passed through 8 further minima, stopping on the 8th. The distance moved is $11.2 \pm 0.2\ \mathrm{cm}$.

(a) Explain how the standing wave is formed. [1]

(b) Determine the frequency of the microwaves. [3]

(d) The readings at the minima are not zero. Suggest why. [1]

Microwaves reflected by the plate superpose with the incident microwaves (travelling in the opposite direction)(1)
Adjacent minima (nodes) are $\lambda/2$ apart, so $8 \times \dfrac{\lambda}{2} = 11.2\ \mathrm{cm}$(1)
$\lambda = 2.8\ \mathrm{cm}$ (0.028 m)(1)
$f = \dfrac{c}{\lambda} = \dfrac{3.00\times10^{8}}{0.028} = 1.07\times10^{10}\ \mathrm{Hz}$ (accept $1.1\times10^{10}$ Hz; ECF for wrong $\lambda$, e.g. $9\lambda/2$ or $8\lambda$ loses the first two marks in (b) only)(1)
Fractional uncertainty $= 0.2/11.2 = 1.8\%$, so $\Delta f \approx 0.02\times10^{10}\ \mathrm{Hz}$ (accept $2\times10^{8}$ Hz; allow $1.9\times10^{8}$ Hz)(1)
The reflected wave has a smaller amplitude than the incident wave (energy absorbed/scattered, wave spreads out), so cancellation at the nodes is incomplete (accept: detector has finite size / averages over a region)(1)
Answer: (b) $1.07\times10^{10}\ \mathrm{Hz}$ (c) $\pm 0.02\times10^{10}\ \mathrm{Hz}$
Examiner tip: Count intervals, not minima: moving from one minimum through 8 more covers 8 half-wavelengths.

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