C.4 Standing waves and resonance: IB Physics SL exam-style questions
A standing wave forms when two identical waves travelling in opposite directions superpose. You need nodes and antinodes, the relative amplitude and phase of points along the wave, and the harmonics on strings and in pipes with open and closed ends, including the wavelength and frequency of the nth harmonic.
Resonance happens when the driving frequency matches the natural frequency of a system. You need the effect of damping on the maximum amplitude and the resonant frequency, light, critical and heavy damping, and the useful and destructive effects of resonance. Pipes are described with displacement nodes only.
25 questions
165 marks
Paper 1A: 11
Paper 1B: 6
Paper 2: 8
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25 practice questions on C.4 Standing waves and resonance
1C-1B-05
Standing waves in pipes·C.4 Standing waves and resonance
Paper 1BEasy8 marks
Data-based question8 steps to full marksDetermine
A narrow plastic tube is closed at one end by a piston that can slide along the tube, so that the length L of the air column can be changed. A small loudspeaker connected to a signal generator is placed at the open end. For each length the student increases the frequency from a low value and records the frequency f at which the sound first becomes loudest. This is the first harmonic of the air column. The uncertainty in f is ±3 Hz. The air temperature is 22 °C.
L / m ± 0.002 m
(1/L) / m−1
f / Hz ± 3 Hz
0.200
5.00
428
0.250
4.00
345
0.300
287
0.400
2.50
217
0.500
2.00
173
0.600
1.67
143
Graph drawn to scale
(a)
State where the displacement node and the displacement antinode are for the first harmonic.
(1)
(b)
Complete the table for L = 0.300 m.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the speed of sound in the tube using the graph.
(3)
(e)
The accepted speed of sound at 22 °C is 344 m s−1. Calculate the percentage difference between the student's value and the accepted value.
(1)
(f)
Suggest one improvement to the method of finding f.
(1)
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Part (a)
Node at the piston (closed end) AND antinode at the open end
✓ 1
Both needed.
Part (b)
3.33 m−1
✓ 1
Accept 3.3 m−1.
Part (c)
Single straight line passing close to the origin with points on both sides
✓ 1
Part (d)
gradient = 85.4 Hz m (m s−1)
✓ 1
Accept 84 to 88. Must use a large triangle.
For a closed tube L = λ/4, so f = v/(4L) and gradient = v/4
✓ 1
v = 4 × 85.4 = 342 m s−1
✓ 1
Accept 336 to 352 m s−1. Allow ECF from the gradient.
Part (e)
|342 − 344|/344 × 100 = 0.6 %
✓ 1
Allow ECF from (d).
Part (f)
Use a microphone connected to an oscilloscope/data-logger to find the frequency of maximum amplitude, OR find the resonance approaching from above and from below and take the mean
✓ 1
Judging 'loudest' by ear is subjective. Do not accept 'repeat' alone.
Answers: (b) 3.33 m−1 · (d) v ≈ 342 m s−1 · (e) ≈ 0.6 % (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — standing waves patterns in strings and pipes; the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions Command term: Determine
2C-1B-06
Tension & wave speed·C.4 Standing waves and resonance
Paper 1BHard12 marks
Data-based question12 steps to full marksDetermine
A steel wire passes over two bridges (0.600 ± 0.001) m apart and then over a pulley to a hanger carrying masses m. The tension T in the wire is taken to be mg. The wire is plucked at its centre and a phone app measures the frequency f of the first harmonic.
The speed of transverse waves on the wire is v = √(T/μ), where μ is the mass per unit length. The student's hypothesis is that f is proportional to √T.
m / kg
T / N
f / Hz
log( T / N )
log( f / Hz )
1.00
9.81
83.6
0.992
1.922
2.00
19.6
117.4
1.293
2.070
3.00
29.4
143.8
1.469
2.158
4.00
166.3
5.00
49.1
185.9
1.691
2.269
6.00
58.9
203.7
1.770
2.309
Graph drawn to scale
(a)
Complete the table for m = 4.00 kg.
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line and state whether the data support the hypothesis.
(3)
(d)
Show that log f = ½ log T − log(2L√μ), where L is the distance between the bridges, and hence determine μ using the graph.
(3)
(e)
A 2.000 m sample of the same wire, measured with an uncertainty of ±0.002 m, has a mass of (1.97 ± 0.01) g. Determine μ and its absolute uncertainty from these data.
(2)
(f)
Friction at the pulley means that the tension in the wire is slightly less than mg. State and explain the effect of this on the value of μ found from the graph.
(1)
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Notes
Part (a)
T = 4.00 × 9.81 = 39.2 N AND log( T / N ) = 1.594
✓ 1
Accept 2 or 4 decimal places. Accept 1.593 (from T = 39.2 N).
log( f / Hz ) = log 166.3 = 2.221
✓ 1
Part (b)
Single straight line with points evenly distributed on both sides
✓ 1
Part (c)
Uses a triangle covering more than half the line
✓ 1
gradient = 0.50
✓ 1
Accept 0.48 to 0.52.
A gradient of 0.5 on a log–log graph means f ∝ T0.5, so the hypothesis is supported
✓ 1
Allow ECF from the gradient.
Part (d)
f = v/(2L) = (1/(2L))√(T/μ); taking logs gives the result
✓ 1
Must see f = v/2L for the first harmonic.
Uses a point on the line, e.g. log( T / N ) = 1.40 and log( f / Hz ) = 2.124: log(2L√μ) = 0.700 − 2.124 = −1.424
✓ 1
Accept the candidate's own point on the line.
2L√μ = 0.0377, so μ = 9.9 × 10−4 kg m−1
✓ 1
Accept 9.6 × 10−4 to 1.0 × 10−3 kg m−1. Allow ECF.
Part (e)
μ = 1.97 × 10−3/2.000 = 9.85 × 10−4 kg m−1
✓ 1
Δμ/μ = 0.01/1.97 + 0.002/2.000 = 0.0061, so μ = (9.85 ± 0.06) × 10−4 kg m−1
The tension used in the analysis (mg) is too large, and μ = T/(4L2f2), so the value of μ from the graph is too large (overestimated)
✓ 1
Direction must be correct and justified.
Answers: (a) 39.2 N; 1.594; 2.221 · (c) 0.50 · (d) μ ≈ 9.9 × 10−4 kg m−1 · (e) μ = (9.85 ± 0.06) × 10−4 kg m−1(the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — standing waves patterns in strings and pipes; nodes and antinodes, relative amplitude and phase difference of points along a standing wave; Tool 3 — linearizing a power law with a log–log graph Command term: Determine
3C-1B-09
Standing waves on strings·C.4 Standing waves and resonance
Paper 1BEasy8 marks
Data-based question8 steps to full marksDetermine
One end of a string is attached to a vibration generator and the other passes over a pulley to a hanging mass. The vibrating length between the generator and the pulley is (1.500 ± 0.005) m. The frequency of the generator is slowly increased, and the frequency f is recorded each time a large-amplitude standing wave with n loops forms. Each frequency has an uncertainty of ±0.5 Hz.
number of loops n
f / Hz ± 0.5 Hz
1
15.8
2
32.2
3
48.1
4
64.1
5
79.7
6
96.0
Graph drawn to scale
(a)
Show that the wavelength of the standing wave with three loops is 1.00 m.
(1)
(b)
State the phase difference between a point in one loop and a point in the adjacent loop.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line of best fit.
(2)
(e)
Determine the speed of the waves on the string.
(2)
(f)
Suggest one way of judging more precisely the frequency at which each standing wave forms.
(1)
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Notes
Part (a)
Three loops contain 3 half-wavelengths, so λ = 2L/3 = 2 × 1.500/3 = 1.00 m
✓ 1
Must see the reasoning.
Part (b)
π rad (180°)
✓ 1
Accept 'antiphase'. Do not accept π/2.
Part (c)
Single straight line passing close to the origin with points on both sides
✓ 1
Part (d)
Uses a triangle covering more than half the line
✓ 1
gradient = 16.0 Hz
✓ 1
Accept 15.7 to 16.3 Hz.
Part (e)
f = nv/(2L), so gradient = v/(2L)
✓ 1
v = 2 × 1.500 × 16.0 = 48.0 m s−1
✓ 1
Accept 47 to 49 m s−1. Allow ECF from (d). Award [2] for CNA.
Part (f)
Approach each resonance from both a lower and a higher frequency and take the mean, OR use a stroboscope/video to find the frequency of maximum amplitude
✓ 1
Do not accept 'repeat' without a method.
Answers: (a) 1.00 m · (b) π rad · (d) 16.0 Hz · (e) v ≈ 48.0 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — standing waves patterns in strings and pipes; nodes and antinodes, relative amplitude and phase difference of points along a standing wave Command term: Determine
4C-1B-10
Resonance & damping·C.4 Standing waves and resonance
Paper 1BMedium9 marks
Data-based question9 steps to full marksDetermine
A trolley of mass (0.500 ± 0.001) kg is held between two springs on a horizontal track. The combined spring constant, measured in a separate experiment, is (40.0 ± 1.0) N m−1. The far end of one spring is moved back and forth by a motor at a driving frequency f. When the motion is steady, a motion sensor records the amplitude A of the trolley. Each amplitude has an uncertainty of ±0.2 cm.
f / Hz
A / cm ± 0.2 cm
0.60
1.2
0.80
1.4
1.00
1.9
1.20
3.4
1.40
7.8
1.60
3.3
1.80
1.6
2.00
1.0
Graph drawn to scale
(a)
Draw the curve of best fit on the graph.
(1)
(b)
Estimate the resonant frequency from the graph.
(1)
(c)
Determine the natural frequency of the trolley–spring system and its absolute uncertainty, using the values of mass and spring constant.
(3)
(d)
Comment on your answers to (b) and (c).
(1)
(e)
Suggest an improvement to the procedure that would allow the resonant frequency to be found more precisely.
(1)
(f)
A large piece of card is fixed to the trolley so that air resistance increases. Describe two changes to the graph of A against f.
(2)
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Notes
Part (a)
Smooth curve with a single peak between 1.4 Hz and 1.5 Hz, rising above the highest point, and falling on both sides through the error bars
✓ 1
Do not accept dot-to-dot.
Part (b)
1.4 Hz
✓ 1
Accept 1.38 to 1.48 Hz.
Part (c)
f0 = (1/2π)√(k/m) = (1/2π)√(40.0/0.500) = 1.42 Hz
✓ 1
Uses T = 2π√(m/k) and f = 1/T.
Δf/f = ½(Δk/k + Δm/m) = ½(0.025 + 0.002) = 0.014
✓ 1
The factor of ½ from the square root is needed.
f0 = (1.42 ± 0.02) Hz
✓ 1
MP3 is for matching the precision of value and uncertainty.
Part (d)
The resonant frequency from the graph lies within the range 1.40–1.44 Hz, so resonance occurs at (close to) the natural frequency
✓ 1
Allow ECF from (b) and (c).
Part (e)
Take more readings at smaller intervals of driving frequency between about 1.2 Hz and 1.6 Hz (close to the peak)
✓ 1
Must be specific to the region near the peak.
Part (f)
The maximum amplitude is smaller (peak lower)
✓ 1
Accept 'amplitude smaller at all frequencies, especially near resonance'.
The peak is broader OR the resonant frequency is slightly lower
✓ 1
Either for the second mark.
Answers: (b) ≈ 1.4 Hz · (c) f0 = (1.42 ± 0.02) Hz (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the effect of damping on the maximum amplitude and resonant frequency of oscillation; C.1 — the time period of a mass–spring system as given by T = 2π√(m/k) Command term: Determine
5C-2-11
Standing waves on strings·C.4 Standing waves and resonance
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksExplain
One end of a rubber cord is attached to a vibration generator G. The generator moves the cord with a very small amplitude, so this end behaves as a fixed end. The other end of the cord is tied to a light ring that can slide without friction along a vertical rod, so this end is free to move up and down.
The length of the cord is 1.50 m and waves travel along it at 9.0 m s−1.
Figure 1 — Diagram NOT to scale vertically
(a)
Explain how a standing wave forms on the cord.
(2)
(b)
(i)
On Figure 1(a), draw the shape of the cord at an instant of maximum displacement when it vibrates in its first harmonic.
(1)
(ii)
Show that the frequency of the first harmonic is 1.5 Hz.
(1)
(c)
The frequency of the generator is slowly increased from 1.5 Hz. Explain why no standing wave forms on the cord at a frequency of 3.0 Hz.
(3)
(d)
(i)
State the frequency of the next standing wave above 1.5 Hz.
(1)
(ii)
On Figure 1(b), draw the shape of the cord at an instant of maximum displacement for this standing wave.
(2)
(iii)
Points P and Q on the cord are 0.40 m and 1.30 m from the generator. State and explain the phase difference between the oscillations of P and Q in this standing wave.
(2)
(e)
The ring is now clamped so that this end of the cord is also fixed. Calculate the frequency of the new first harmonic.
(1)
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Notes
Part (a)
waves travelling along the cord are reflected at the ends
✓ 1
the incident and reflected waves «same frequency, speed and similar amplitude», travelling in opposite directions, superpose
✓ 1
OWTTE
Part (b)(i)
curve with a node at G and an antinode at the ring, and no other node «a quarter of a wavelength»
✓ 1
Part (b)(ii)
λ = 4 × 1.50 = 6.0 m and f = 9.0/6.0 «= 1.5 Hz»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
at 3.0 Hz the wavelength is 9.0/3.0 = 3.0 m, so the cord would be half a wavelength long
✓ 1
a half-wavelength pattern needs the same type of point «node or antinode» at both ends
✓ 1
but there must be a node at G and an antinode at the ring, so the length must be an odd number of quarter wavelengths «frequencies 1.5, 4.5, 7.5 … Hz»
✓ 1
OWTTE
Part (d)(i)
4.5 Hz
✓ 1
Part (d)(ii)
nodes at G and at 1.00 m from G
✓ 1
Allow ECF from (d)(i)
antinodes at 0.50 m and at the ring «three quarters of a wavelength»
✓ 1
Part (d)(iii)
π rad «180°, antiphase»
✓ 1
P and Q are in adjacent loops, on opposite sides of the node at 1.00 m
✓ 1
MP2 only scores if MP1 scores
Part (e)
λ = 2 × 1.50 = 3.0 m, so f = 9.0/3.0 = 3.0 Hz
✓ 1
Answers: (b)(ii) 1.5 Hz · (d)(i) 4.5 Hz · (d)(iii) π rad · (e) 3.0 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions; nodes and antinodes, relative amplitude and phase difference of points along a standing wave; standing waves patterns in strings Command term: Explain
6C-2-12
Standing waves in pipes·C.4 Standing waves and resonance
Paper 2Medium11 marks
Short answer & extended response11 steps to full marksDetermine whether
A simple wind instrument is modelled as a straight pipe of length 0.660 m that is open at both ends. The speed of sound in the air in the pipe is 343 m s−1.
Figure 1 — Diagram NOT to scale
(a)
State whether there is a displacement node or a displacement antinode at an open end of the pipe.
(1)
(b)
Sketch, on Figure 1(a), the displacement pattern of the first harmonic of the air in the pipe.
(2)
(c)
Show that the frequency of the first harmonic is about 260 Hz.
(1)
(d)
Figure 1(b) shows the displacement pattern of another standing wave in the pipe. Determine its frequency.
(2)
(e)
The player closes one end of the pipe. Determine whether the pipe can now form a standing wave of the frequency in (d).
(3)
(f)
The air in the pipe warms up while the instrument is played. Suggest, with a reason, the effect on the frequency of the first harmonic.
(2)
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Notes
Part (a)
a displacement antinode
✓ 1
Part (b)
displacement antinodes at both open ends
✓ 1
a single node at the centre of the pipe «half a wavelength in the pipe»
✓ 1
Do not award MP2 for a quarter-wavelength pattern
Part (c)
λ = 2 × 0.660 = 1.32 m and f = 343/1.32 «= 259.8 Hz»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (d)
three half-wavelengths fit in the pipe: λ = 2 × 0.660/3 = 0.440 m
✓ 1
Accept f = 3f1
f = 343/0.440 = 780 Hz
✓ 1
Allow ECF from (c). Award [2] for CNA
Part (e)
with one end closed, standing waves form only at odd multiples of v/4L
✓ 1
v/4L = 343/(4 × 0.660) = 130 Hz
✓ 1
780/130 = 6, an even number, so this standing wave cannot form
✓ 1
MP3 only scores if a conclusion is given. Allow ECF from (d)
Part (f)
the speed of sound increases with temperature
✓ 1
the wavelength is fixed by the length of the pipe, so f = v/λ increases
✓ 1
MP2 only scores if MP1 scores. OWTTE
Answers: (c) 259.8 Hz · (d) 780 Hz · (e) no: 780 Hz = 6 × 130 Hz (even multiple) (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — standing waves patterns in strings and pipes; nodes and antinodes (boundary conditions: two open ends, and one closed and one open end) Command term: Determine whether
7C-2-13
Resonance & damping·C.4 Standing waves and resonance
Paper 2Medium14 marks
Short answer & extended response14 steps to full marksSketch
A kitchen scale has a pan of mass 0.200 kg supported on a vertical spring of spring constant 320 N m−1. A damper is connected to the pan. Just before time t = 0 the pan is at rest at its original position. At t = 0 a bag of flour of mass 0.300 kg is placed on the pan and released from rest. A sensor records the downward displacement y of the pan from its original position.
Figure 1 shows the variation of y with t for the scale with its original damper. Take g = 9.81 m s−2.
Figure 1 — drawn to scale
(a)
Outline what is meant by damping.
(1)
(b)
Show that the final position of the pan is about 9 mm below its original position.
(1)
(c)
Describe how Figure 1 shows that the motion of the pan is lightly damped.
(2)
(d)
Sketch, on Figure 1, graphs to show how y would vary with t if the damping were critical (label this graph C) and if the damping were heavy (label this graph H).
(3)
(e)
Suggest why the manufacturer designs the scale to be critically damped.
(2)
(f)
(i)
Determine the total energy transferred to thermal energy by the damping between t = 0 and the time at which the pan comes to rest.
(3)
(ii)
State and explain whether this energy would be greater, smaller or the same if the scale were critically damped.
(2)
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Notes
Part (a)
a resistive force «friction, drag» that acts on an oscillating system, opposite to its velocity, and removes energy from it so that the amplitude decreases
✓ 1
OWTTE
Part (b)
extra spring force = weight of the bag: d = Mg/k = 0.300 × 9.81/320 «= 9.20 mm»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
the pan oscillates about its final position «9.2 mm», overshooting it several times
✓ 1
the amplitude decreases gradually «by a similar fraction each cycle» while the period stays the same «about 0.25 s»
✓ 1
OWTTE
Part (d)
both C and H start at y = 0 with zero gradient and rise to 9.2 mm without ever exceeding it «no oscillation»
✓ 1
Labels C and H required
C reaches «close to» 9.2 mm quickly, in about one period of the light oscillation «≈ 0.2 s»
✓ 1
Accept a time from 0.1 s to 0.4 s
H lies below C at all times after t = 0 and approaches 9.2 mm much more slowly
✓ 1
MP3 only scores if both curves are labelled
Part (e)
with light damping the pan oscillates for a long time «about 1 s in Figure 1» before the reading is steady / with heavy damping the reading creeps slowly towards its final value
✓ 1
OWTTE
critical damping brings the pan to its final reading, without oscillation, in the shortest time
✓ 1
Part (f)(i)
loss of gravitational potential energy of pan and bag = 0.500 × 9.81 × 0.00920 = 0.0451 J
✓ 1
Allow ECF from (b)
gain of elastic potential energy = ½ × 320 × (0.01533² − 0.006131²) = 0.0316 J «initial compression 6.13 mm, final 15.3 mm»
✓ 1
ALTERNATIVE: the pan's weight and the initial spring force cancel, so energy = Mgd − ½kd² = 0.0271 − 0.0135 = 0.0135 J
thermal energy = 0.0451 − 0.0316 = 0.0135 J
✓ 1
Accept 0.013–0.014 J. Award [3] for CNA. Award [1] max for Mgd = 0.0271 J
Part (f)(ii)
the same
✓ 1
Do not award MP1 without a reason
the initial and final states are the same «at rest, same positions, so the same gravitational and elastic potential energies», so by conservation of energy the same energy is dissipated; only the time taken is different
✓ 1
OWTTE
Answers: (b) d = 9.20 mm · (f)(i) 0.0135 J · (f)(ii) the same (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — the effects of light, critical and heavy damping on the system (with A.3 — the principle of the conservation of energy; the elastic potential energy stored in a spring as given by EH = ½k(Δx)²) Command term: Sketch
8C-1A-29
Standing waves on strings·C.4 Standing waves and resonance
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A string of length 0.65 m is fixed at both ends. Waves travel along the string at 260 m s−1.
What is the frequency of the third harmonic?
Diagram NOT accurately drawnShow mark scheme
Marking point
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Notes
Step 1The first harmonic has half a wavelength on the string: f1 = v/2L = 260/1.30 = 200 Hz.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The third harmonic has three times this frequency: f3 = 3 × 200 = 600 Hz.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis takes the first-harmonic wavelength as L instead of 2L, giving 400 Hz, and then multiplies by 3.
BCorrect: f1 = v/2L = 200 Hz, so f3 = 3f1 = 600 Hz.
CThis is the second harmonic, 2 × 200 Hz.
DThis is the first harmonic; the third harmonic has three times its frequency.
Syllabus understandingC.4 — standing waves patterns in strings and pipes Command term: Determine
9C-1A-30
Standing waves in pipes·C.4 Standing waves and resonance
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A pipe of length L is closed at one end and open at the other. The speed of sound in the air in the pipe is v.
What is the frequency of the standing wave with the lowest frequency above that of the first harmonic?
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Notes
Step 1There is a displacement node at the closed end and an antinode at the open end, so the pipe holds an odd number of quarter-wavelengths: f1 = v/4L, and only odd multiples of f1 occur.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The next standing wave is therefore the third harmonic, 3v/4L.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is the first harmonic itself, not the next standing wave above it.
BThis is 2f1. A pipe closed at one end cannot have an antinode at the closed end, so even harmonics do not occur.
CCorrect: only odd harmonics fit a pipe closed at one end, so the next one above v/4L is 3v/4L.
DThis is the second harmonic of a pipe open at both ends, 2 × v/2L; here one end is closed.
Syllabus understandingC.4 — standing waves patterns in strings and pipes Command term: Determine
10C-1A-31
Standing waves in pipes·C.4 Standing waves and resonance
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A pipe open at both ends has a first harmonic of frequency fopen. One end of the pipe is then closed, and the new first harmonic has frequency fclosed. The speed of sound is unchanged.
What is fclosed/fopen?
Show mark scheme
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Mark
Notes
Step 1Open at both ends: antinodes at both ends, so λ1 = 2L. Closed at one end: node to antinode, so λ1 = 4L.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2f = v/λ, so fclosed/fopen = 2L/4L = 1/2.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis takes the open pipe's first harmonic as one whole wavelength (λ = L); from antinode to antinode is only half a wavelength.
BCorrect: the first-harmonic wavelength doubles from 2L to 4L, so the frequency halves.
CThe length is unchanged, but the boundary condition at the closed end changes which wavelengths fit.
DThis is the inverse ratio. Closing one end lengthens the first-harmonic wavelength, which lowers the frequency.
Syllabus understandingC.4 — standing waves patterns in strings and pipes Command term: Determine
11C-1A-32
Nodes & antinodes·C.4 Standing waves and resonance
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A standing wave is set up on a string fixed at both ends. The diagram shows the string at one instant of maximum displacement (solid line) and half a period later (dashed line). P, Q and R are points on the string. The distance PQ is λ/4 and the distance PR is 3λ/4, where λ is the wavelength.
Which row gives the phase difference between the oscillations of P and Q, and between those of P and R?
Diagram drawn to scale
Between P and QBetween P and R
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Marking point
Mark
Notes
Step 1P and Q lie between the same pair of adjacent nodes, so they oscillate in phase: phase difference 0.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2R lies in the neighbouring loop, on the other side of a node, so it oscillates in antiphase with P: phase difference π.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis applies the travelling-wave rule, phase difference = 2π × separation/λ, to both pairs; in a standing wave the phase does not change steadily with position.
BCorrect: points in the same loop are in phase, and points in adjacent loops are in antiphase.
CP and R are right, but P and Q have been given the travelling-wave phase difference for points λ/4 apart; they are in the same loop, so they are in phase.
DP and Q are right, but P and R have been given the travelling-wave phase difference for points 3λ/4 apart; one node separates them, so they are in antiphase.
Syllabus understandingC.4 — nodes and antinodes, relative amplitude and phase difference of points along a standing wave Command term: Deduce
12C-1A-33
Standing waves on strings·C.4 Standing waves and resonance
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
A string is clamped at end X. Its other end, Y, is tied to a light ring that slides without friction on a vertical rod, so that Y is a free end.
Which diagram shows the third harmonic of the string? Each diagram shows the two extreme positions of the string.
Diagrams NOT accurately drawnShow mark scheme
Marking point
Mark
Notes
Step 1The clamped end X is a node and the free end Y is an antinode, so the string holds an odd number of quarter-wavelengths; the first harmonic is a quarter-wavelength.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The third harmonic has three times the first-harmonic frequency, so three quarter-wavelengths fit on the string: node at X, one node between, antinode at Y.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: a node at X, an antinode at Y and three quarter-wavelengths on the string, so the frequency is three times that of the first harmonic.
BThe boundary conditions are reversed: this has an antinode at the clamped end X and a node at the free end Y.
CThis has five quarter-wavelengths on the string: it is the third possible standing wave, but its frequency is 5 times that of the first harmonic, so it is the fifth harmonic.
DThis is the third harmonic of a string fixed at both ends; it puts a node at Y, but a free end is an antinode.
Syllabus understandingC.4 — standing waves patterns in strings and pipes Command term: Identify
13C-1A-34
Tension & wave speed·C.4 Standing waves and resonance
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
The tension in a string fixed at both ends is increased to four times its original value. The length of the string does not change. The speed of waves on the string is proportional to the square root of the tension.
What is (new first-harmonic frequency)/(original first-harmonic frequency)?
Show mark scheme
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Mark
Notes
Step 1v ∝ √(tension), so the wave speed is multiplied by √4 = 2.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The first-harmonic wavelength, 2L, is unchanged, so f = v/2L is multiplied by 2.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is the inverse ratio: a tighter string carries faster waves, so it has a higher first-harmonic frequency.
BThe wavelength is fixed by the length of the string, but the wave speed changes, so the frequency must change too.
CCorrect: the wave speed doubles and the wavelength of the first harmonic is unchanged, so the frequency doubles.
DThis applies the factor of 4 to the speed directly; the wave speed depends on the square root of the tension.
Syllabus understandingC.4 — standing waves patterns in strings and pipes Command term: Determine
14C-1A-35
Resonance & damping·C.4 Standing waves and resonance
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Three oscillators with different degrees of damping are each displaced by the same amount and released at time t = 0. The graph shows how the displacement of each oscillator varies with time.
Which row identifies the critically damped oscillator and the heavily damped oscillator?
Graph drawn to scale
Critically dampedHeavily damped
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Notes
Step 1Q oscillates about equilibrium with decreasing amplitude, so it is lightly damped; of P and R, which do not oscillate, R returns to equilibrium sooner, so R is critically damped and P is heavily damped.
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AP and R have been swapped: heavy damping slows the return, and critical damping returns to equilibrium in the shortest time without oscillating.
BQ oscillates about the equilibrium position, so it is lightly damped, not critically damped.
CCorrect: R returns to equilibrium in the shortest time without oscillating (critical damping), while P returns more slowly (heavy damping).
DQ oscillates, so it is lightly damped; and R, which returns fastest without oscillating, is critically rather than heavily damped.
Syllabus understandingC.4 — the effects of light, critical and heavy damping on the system Command term: Identify
15C-1A-36
Resonance & damping·C.4 Standing waves and resonance
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDescribe
A system is made to oscillate by a periodic driving force whose frequency can be varied. The damping of the system is then increased.
Which statement describes the change to the graph of amplitude against driving frequency?
Show mark scheme
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Notes
Step 1Heavier damping removes energy from the system faster, so the amplitude at resonance is smaller.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The response is spread over a wider range of driving frequencies (a broader peak), and the frequency of maximum amplitude moves slightly below the natural frequency.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the maximum amplitude falls, the peak broadens, and the resonant frequency moves slightly lower.
BHeavier damping flattens the curve, spreading the response over a wider range of driving frequencies, so the peak becomes broader, not narrower.
CThe damping removes more energy per cycle, so the maximum amplitude must fall.
DThe height and width are right, but the shift is the wrong way: increased damping moves the peak slightly below the natural frequency.
Syllabus understandingC.4 — the effect of damping on the maximum amplitude and resonant frequency of oscillation Command term: Describe
16C-1A-37
Standing vs travelling waves·C.4 Standing waves and resonance
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A standing wave and a travelling wave of the same frequency exist on two identical strings.
Which statement is true of the standing wave but not of the travelling wave?
Show mark scheme
Marking point
Mark
Notes
Step 1In a travelling wave the phase changes steadily with position, so two points less than half a wavelength apart are never in phase.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2In a standing wave every point between two adjacent nodes (half a wavelength apart) oscillates in phase, so such points can be in phase.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is true of the travelling wave; in a standing wave the amplitude varies from zero at a node to a maximum at an antinode.
BThis is true of the travelling wave; a standing wave transfers no net energy along the string.
CCorrect: all points in one loop of a standing wave are in phase, whereas in a travelling wave the phase changes continuously along the wave.
DThis is true of the travelling wave as well (apart from the nodes of the standing wave, which do not move), so it does not distinguish them.
Syllabus understandingC.4 — the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions; nodes and antinodes, relative amplitude and phase difference of points along a standing wave Command term: Deduce
17C-1A-38
Superposition·C.4 Standing waves and resonance
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksDetermine
Two waves that are identical except for their direction of travel move in opposite directions along a string and form a standing wave.
What is (amplitude of the standing wave at an antinode)/(amplitude of each travelling wave)?
Show mark scheme
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Mark
Notes
Step 1At an antinode the two waves always arrive in phase, so their displacements add: the amplitude is 2 × the amplitude of each wave, and the ratio is 2.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThis is the value at a node, where the two waves are always in antiphase and cancel.
BThis assumes the standing wave has the same amplitude as each travelling wave; at an antinode the two waves reinforce each other.
CThis adds the two amplitudes as though they were perpendicular vectors. Superposition adds displacements along the same line.
DCorrect: at an antinode the displacements of the two waves are always equal and in the same direction, so the amplitude doubles.
Syllabus understandingC.4 — the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions Command term: Determine
18C-1A-39
Resonance & damping·C.4 Standing waves and resonance
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksExplain
As the drum of a washing machine speeds up, the whole machine vibrates violently at one particular drum speed. At higher drum speeds the vibration becomes much smaller.
Which statement explains this?
Show mark scheme
Marking point
Mark
Notes
Step 1The rotating drum is a periodic driving force; the amplitude of the forced oscillation is greatest when the driving frequency equals the natural frequency of the machine on its supports — resonance.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: at resonance energy is transferred from the driver to the machine most effectively, so the amplitude is at its largest.
BThe out-of-balance force from the drum grows as the drum speeds up; the large amplitude comes from matching frequencies, not from a larger force.
CThe natural frequency is fixed by the mass and stiffness of the machine and its supports; it is the driving frequency that changes.
DDamping does not disappear at any speed; it limits the amplitude at resonance to a finite value.
Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency Command term: Explain
19C-1B-17
Standing waves in pipes·C.4 Standing waves and resonance
Paper 1BMedium10 marks
Data-based question10 steps to full marksDeduce
A student investigates the air in a plastic pipe of length (0.850 ± 0.002) m. One end of the pipe is open; the student cannot see whether the far end has been sealed. A small loudspeaker at the open end plays a short burst of noise, and a microphone connected to a computer records the sound. The frequency spectrum shows peaks at the resonant frequencies of the air column. The frequencies are listed in order of increasing frequency as peak number n. Each frequency has an uncertainty of ±2 Hz.
peak number n
2n − 1
f / Hz ± 2 Hz
1
1
101
2
3
303
3
5
504
4
7
706
5
9
908
Graph drawn to scale
(a)
Show that the ratio of the frequencies of the second and first peaks is about 3.
(1)
(b)
Deduce, with reference to the data, whether the far end of the pipe is sealed.
(2)
(c)
The graph shows f against (2n − 1). Draw the line of best fit on the graph.
(1)
(d)
Determine the speed of sound in the pipe using the graph.
(3)
(e)
The uncertainty in the gradient is ±0.5 Hz. Determine the absolute uncertainty in the speed of sound.
(2)
(f)
The accepted speed of sound at the temperature of the laboratory is 343 m s−1. Comment on the student's result.
(1)
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Notes
Part (a)
303/101 = 3.00
✓ 1
Must see the division AND an answer to at least 3 s.f.
Part (b)
The frequencies are in the ratio 1 : 3 : 5 : 7 : 9 (only odd multiples of the lowest frequency, 101 Hz)
✓ 1
Must use the data, e.g. two further ratios.
This is the pattern for a pipe closed at one end (an open pipe would give all whole-number multiples 1 : 2 : 3 …), so the far end is sealed
✓ 1
Conclusion must follow from the ratios.
Part (c)
Single straight line through all points and passing close to the origin
✓ 1
Part (d)
gradient = 100.9 Hz
✓ 1
Accept 99 to 103 Hz.
For a pipe closed at one end f = (2n − 1)v/(4L), so gradient = v/(4L)
✓ 1
v = 4 × 0.850 × 100.9 = 343 m s−1
✓ 1
Accept 337 to 350 m s−1. Allow ECF. Award [3] for CNA with the relation seen.
Part (e)
Δv/v = 0.5/100.9 + 0.002/0.850 = 0.0073
✓ 1
Both fractional uncertainties needed.
Δv = ±3 m s−1, so v = (343 ± 3) m s−1
✓ 1
Accept ±2.5 m s−1. MP2 includes matching the precision of value and uncertainty.
Part (f)
343 m s−1 lies within the range v ± Δv, so the result is consistent with the accepted value
✓ 1
Allow ECF from (d) and (e).
Answers: (a) 3.00 · (d) v ≈ 343 m s−1 · (e) Δv ≈ ±3 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — standing waves patterns in strings and pipes; the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions Command term: Deduce
20C-1B-19
Nodes & antinodes·C.4 Standing waves and resonance
Paper 1BEasy8 marks
Data-based question8 steps to full marksDetermine
A microwave transmitter faces a flat metal plate. A small probe receiver is moved along the line between them, and the student records the distance x from the plate of each position where the received signal is a minimum. The minima are numbered N = 1, 2, 3 … starting with the one nearest the plate. Each position has an uncertainty of ±0.2 cm.
N
x / cm ± 0.2 cm
1
1.4
2
2.9
3
4.3
4
5.7
5
7.1
6
8.5
7
10.0
8
11.4
Graph drawn to scale
(a)
Explain how the pattern of maxima and minima is formed.
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the wavelength of the microwaves using the graph.
(2)
(d)
The uncertainty in the gradient is ±2 %. Determine the frequency of the microwaves and its absolute uncertainty.
(2)
(e)
The manufacturer states that the transmitter emits at 10.5 GHz. Comment on your answer to (d).
(1)
Show mark scheme
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Mark
Notes
Part (a)
The microwaves reflect from the plate, and the reflected wave travels back towards the transmitter with the same frequency and speed as the incident wave
✓ 1
Must refer to reflection.
The two waves travelling in opposite directions superpose to form a standing wave: the minima are nodes (destructive superposition) and the maxima are antinodes
✓ 1
OWTTE.
Part (b)
Single straight line through all error bars, passing close to the origin
✓ 1
The plate is a node at x = 0.
Part (c)
gradient = 1.42 cm = distance between adjacent nodes = λ/2
✓ 1
Accept 1.40 to 1.46 cm.
λ = 2 × 1.42 = 2.85 cm
✓ 1
Accept 2.80 to 2.92 cm. Allow ECF. Award [2] for CNA.
Fractional uncertainty in f equals that in λ. Matching precision required.
Part (e)
10.5 GHz lies within the range f ± Δf, so the result is consistent with the manufacturer's value
✓ 1
Allow ECF from (d).
Answers: (c) λ ≈ 2.85 cm · (d) f = (1.05 ± 0.02) × 1010 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions; nodes and antinodes; C.2 — the nature of electromagnetic waves Command term: Determine
21C-2-29
Tension & wave speed·C.4 Standing waves and resonance
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksExplain
An overhead power cable is fixed to one tower and passes over a pulley on a second tower 300 m away. Its mass per unit length is 1.50 kg m−1. The tension in the cable is set by a counterweight of mass 2.20 × 103 kg that hangs from the end of the cable beyond the pulley. Treat the cable between the towers as straight; both of its ends there behave as fixed ends.
The speed v of transverse waves on the cable is given by v = √(T/μ), where T is the tension and μ is the mass per unit length.
Wind blowing across the cable exerts a periodic vertical force on it. For the wind speeds usually met, the frequency of this force lies between 5 Hz and 40 Hz.
(a)
(i)
Show that the speed of transverse waves on the cable is about 120 m s−1.
(2)
(ii)
Calculate the frequency of the first harmonic of the cable.
(1)
(b)
On one day the frequency of the force from the wind is 12.0 Hz and a standing wave is set up on the cable.
(i)
Determine the number of the harmonic that is set up.
(2)
(ii)
Calculate the distance between adjacent nodes of this standing wave.
(1)
(c)
Explain why the cable vibrates with a large amplitude when the force has a frequency of 12.0 Hz, and why a vibration damper clamped to the cable at a node of this standing wave would not reduce the vibration.
(4)
(d)
Suggest why standing waves of large amplitude can occur at almost any wind speed.
(1)
(e)
Vibration dampers are usually clamped to the cable about 1 m from a tower. Explain, using the range of frequencies of the force, why this position is chosen.
(2)
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Notes
Part (a)(i)
Tension = weight of the counterweight: T = mg = 2.20 × 103 × 9.81 = 2.16 × 104 N
✓ 1
v = √(2.158 × 104/1.50) = 119.9 «m s−1»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
f1 = v/2L = 120/600 = 0.200 Hz
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
The harmonics are whole-number multiples of f1, so n = 12.0/0.200
✓ 1
Allow ECF from (a)(ii).
n = 60
✓ 1
Award [2] for CNA.
Part (b)(ii)
λ/2 = L/60 = 300/60 = 5.0 m
✓ 1
Accept v/2f = 120/24.0 = 5.0 m.
Part (c)
The driving frequency «12.0 Hz» equals a natural frequency of the cable «its 60th harmonic»
✓ 1
so resonance occurs: energy is transferred to the cable on every cycle and the amplitude becomes large
✓ 1
A node is a point of zero amplitude — the cable there does not move
✓ 1
so a damper at a node has no motion to oppose and cannot remove energy from the standing wave
✓ 1
OWTTE.
Part (d)
The natural frequencies are only 0.2 Hz apart, so the frequency of the force is always very close to one of them
✓ 1
Allow ECF from (a)(ii).
Part (e)
The first node from the tower is half a wavelength away; at the highest frequency, 40 Hz, λ/2 = 120/(2 × 40) = 1.5 m
✓ 1
So a damper less than 1.5 m from the tower is never at a node for any frequency in the range «it always moves and removes energy»
✓ 1
Award [1 max] for 'it must not be at a node' without the calculation.
Answers: (a)(i) T = 2.16 × 104 N; v = 120 m s−1 · (a)(ii) 0.200 Hz · (b)(i) n = 60 · (b)(ii) 5.0 m · (e) 1.5 m (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — standing waves patterns in strings and pipes; the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; A.2 — weight as given by Fg = mg Command term: Explain
22C-2-30
Standing waves in pipes·C.4 Standing waves and resonance
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksSketch
The ear canal of an adult is a tube about 2.5 cm long. It is open at the outer end and closed at the inner end by the eardrum. The air in the canal can be modelled as the air in a pipe with one open end and one closed end. The speed of sound in the air in the canal is 340 m s−1.
Diagram drawn to scale
(a)
State where there is a displacement node and where there is a displacement antinode in every standing wave in the canal, and explain why.
(2)
(b)
On the upper diagram, sketch the displacement pattern of the standing wave of the first harmonic.
(1)
(c)
Show that the frequency of the first harmonic is about 3.4 kHz.
(1)
(d)
Determine the frequency of the next standing wave above the first harmonic.
(2)
(e)
On the lower diagram, sketch the displacement pattern of the standing wave in (d).
(2)
(f)
Hearing tests show that the human ear is most sensitive to sounds of frequency about 3 kHz to 4 kHz. Suggest an explanation for this.
(2)
(g)
The ear canal of a young child is shorter than that of an adult. Outline how this affects the frequency to which the child's ear is most sensitive.
(1)
Show mark scheme
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Mark
Notes
Part (a)
A displacement node at the eardrum «closed end», where the air cannot move
✓ 1
A displacement antinode at the open end, where the air is free to move
✓ 1
Award [1 max] for the positions without reasons.
Part (b)
Node at the eardrum and antinode at the open end, with no other node: a quarter of a wavelength
✓ 1
Accept the envelope drawn as one curve or as a pair of curves.
Part (c)
λ = 4L = 0.10 m, so f = 340/0.10 = 3.4 × 103 «Hz»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (d)
Only odd harmonics can form: the next has three quarters of a wavelength in the canal
✓ 1
f = 3 × 3.4 = 1.02 × 104 Hz «10.2 kHz»
✓ 1
Do not accept 6.8 kHz. Allow ECF from (c). Award [2] for CNA.
Part (e)
Node at the eardrum and antinode at the open end
✓ 1
One further node one third of the length from the open end «0.83 cm», and one further antinode two thirds of the length from the open end «1.7 cm», between this node and the eardrum
✓ 1
Positions within ±2 mm on the scale of the diagram. Allow ECF from (d).
Part (f)
Sound at about 3.4 kHz matches the natural frequency of the air in the canal «its first harmonic»
✓ 1
so resonance occurs: the air in the canal oscillates with a large amplitude, giving a large amplitude at the eardrum
✓ 1
OWTTE.
Part (g)
The child's canal has a higher first-harmonic frequency «f = v/4L», so the most sensitive frequency is higher
✓ 1
Answers: (c) 3.4 × 103 Hz · (d) 1.02 × 104 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — standing waves patterns in strings and pipes; nodes and antinodes, relative amplitude and phase difference of points along a standing wave; the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency Command term: Sketch
23C-2-31
Resonance & damping·C.4 Standing waves and resonance
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksDetermine
Four simple pendulums P, Q, R and S hang from the same horizontal rod. Their lengths are 0.25 m, 0.39 m, 0.60 m and 0.95 m respectively. Each bob is a light paper cone, so air resistance damps the motion. A motor makes the rod oscillate horizontally with a small amplitude, at a frequency that can be varied.
(a)
The motor drives the rod at a frequency of 0.80 Hz. Determine which pendulum oscillates with the largest amplitude.
(3)
(b)
State the frequency at which pendulum S oscillates once its motion has become steady.
(1)
(c)
Explain why the amplitude of pendulum S is small.
(2)
(d)
The frequency of the motor is now increased slowly from 0.40 Hz to 1.20 Hz. Describe and explain what is observed.
(2)
(e)
The paper cone of pendulum Q is replaced by a small lead sphere, so the damping of Q is lighter. State and explain the effect on the amplitude of Q when the rod is driven at 0.80 Hz and when it is driven at 0.60 Hz.
(2)
Show mark scheme
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Notes
Part (a)
Natural frequency f = (1/2π)√(g/l) OR length needed l = g/(2πf)²
✓ 1
l = 9.81/(2π × 0.80)² = 0.39 m «OR f of Q = 0.80 Hz»
✓ 1
ALTERNATIVE: natural frequencies of P, Q, R, S = 1.00, 0.80, 0.64, 0.51 Hz.
Q, because its natural frequency equals the driving frequency, so resonance occurs
✓ 1
MP3 only scores if a length or frequency has been calculated.
Part (b)
0.80 Hz «the driving frequency»
✓ 1
Do not accept its natural frequency of 0.51 Hz.
Part (c)
The driving frequency is far from the natural frequency of S «0.51 Hz»
✓ 1
so the driving force is not in step with the motion of S: energy given to S during part of each cycle is taken back during another part, and little energy builds up
✓ 1
OWTTE.
Part (d)
Each pendulum in turn oscillates with a large amplitude: S, then R, then Q, then P
✓ 1
because the natural frequency increases as the length decreases «f ∝ 1/√l», and each pendulum resonates when the driving frequency passes its natural frequency
✓ 1
OWTTE.
Part (e)
At 0.80 Hz «resonance» the amplitude of Q is much larger, because lighter damping removes less energy per cycle
✓ 1
At 0.60 Hz, away from resonance, the amplitude changes little «the resonance peak is sharper»
✓ 1
OWTTE.
Answers: (a) l = 0.39 m — pendulum Q · (b) 0.80 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the effect of damping on the maximum amplitude and resonant frequency of oscillation; C.1 — the time period of a simple pendulum as given by T = 2π√(l/g) Command term: Determine
24C-2-32
Standing waves on strings·C.4 Standing waves and resonance
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksExplain
A violin string is fixed at both ends and has a vibrating length of 0.330 m. When it is bowed, the frequency of its first harmonic is 440 Hz.
A violinist can produce higher notes by touching the string lightly at one point while bowing it. The light touch forces a node at that point but does not change the vibrating length of the string.
Diagram NOT accurately drawn
(a)
(i)
Calculate the speed of transverse waves on the string.
(2)
(ii)
The violinist now presses the string firmly against the fingerboard, so that its vibrating length becomes 0.220 m. Calculate the new frequency of the first harmonic. In the rest of the question the string is not pressed.
(2)
(b)
The violinist touches the string lightly at a point one third of the way along it, as shown in the figure.
(i)
On the figure, draw the standing wave of lowest frequency that can now be set up on the string.
(2)
(ii)
Calculate the frequency of this standing wave.
(1)
(iii)
Explain why the first and second harmonics of the string cannot be set up while it is touched at this point.
(2)
(iv)
State the phase difference between the oscillations of points X and Y in this standing wave.
(1)
(c)
The violinist now touches the string lightly at a point one quarter of the way along it. Determine whether a note of frequency 1320 Hz can be produced.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
λ = 2L = 0.660 m
✓ 1
v = fλ = 440 × 0.660 = 290 m s−1
✓ 1
Accept 290.4 m s−1. Award [2] for CNA.
Part (a)(ii)
λ = 2 × 0.220 = 0.440 m
✓ 1
Accept f ∝ 1/L: f = 440 × 0.330/0.220.
f = 290/0.440 = 660 Hz
✓ 1
Allow ECF from (a)(i). Award [2] for CNA.
Part (b)(i)
Nodes at both ends and at the touched point «and at two-thirds of the length»
✓ 1
Three loops of equal length «the third harmonic»
✓ 1
Part (b)(ii)
f = 3 × 440 = 1320 Hz
✓ 1
Part (b)(iii)
The touched point must be a node, but the first harmonic has nodes only at the ends
✓ 1
and the second harmonic has nodes only at the ends and the centre; neither has a node one third of the way along, so neither can exist
✓ 1
OWTTE.
Part (b)(iv)
π rad «180°; X and Y are in antiphase»
✓ 1
X and Y are in adjacent loops.
Part (c)
The point at L/4 must be a node, so only harmonics with a node there can exist: n = 4, 8, …
✓ 1
1320 Hz is the third harmonic, whose nodes are at 0, L/3, 2L/3 and L — not at L/4
✓ 1
So 1320 Hz cannot be produced «the lowest note is now 4 × 440 = 1760 Hz»
✓ 1
MP3 only scores if a reason is given.
Answers: (a)(i) 290 m s−1 · (a)(ii) 660 Hz · (b)(ii) 1320 Hz · (b)(iv) π rad · (c) not possible; lowest note 1760 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions; nodes and antinodes, relative amplitude and phase difference of points along a standing wave; standing waves patterns in strings and pipes Command term: Explain
25C-2-37
Resonance & damping·C.4 Standing waves and resonance
Paper 2Medium14 marks
Short answer & extended response14 steps to full marksExplain
A vibration energy harvester is bolted to a pump in a factory. It consists of a small mass of 12.0 g on the end of a thin steel strip that behaves like a spring, so the mass can oscillate with simple harmonic motion. A magnet on the mass moves inside a coil and generates electrical energy, which removes energy from the oscillation. The pump vibrates at 50.0 Hz and drives the harvester.
The graph shows how the amplitude of oscillation of the mass varies with the driving frequency.
Graph drawn to scale
(a)
Show that the spring constant of the strip must be about 1.2 × 103 N m−1 for the natural frequency of the harvester to be 50.0 Hz.
(1)
(b)
Outline why the harvester is designed to have a natural frequency equal to the frequency of the pump.
(2)
(c)
The same harvester is to be used on a pump that vibrates at 60.0 Hz. Determine the mass that must be removed so that the harvester has this natural frequency.
(2)
(d)
(i)
Estimate, using the graph, the amplitude of the mass when the pump runs at 49.0 Hz.
(1)
(ii)
Suggest one disadvantage of a harvester whose resonance peak is very sharp.
(1)
(e)
At resonance the amplitude of the mass is 4.0 mm.
(i)
Determine the maximum resultant force on the mass.
(3)
(ii)
Suggest why the steel strip may eventually break.
(1)
(f)
Explain, in terms of energy, why the amplitude of the mass at resonance stays constant while the harvester supplies electrical energy.
Must see full substitution OR answer to at least 3 s.f. Accept a rearrangement of T = 2π√(m/k) with T = 0.0200 s.
Part (b)
When the driving frequency equals the natural frequency, resonance occurs
✓ 1
so the amplitude of the mass is a maximum and energy is transferred from the pump to the harvester at the greatest rate «giving the greatest electrical output»
✓ 1
OWTTE.
Part (c)
m = k/(2πf)² = 1184/(2π × 60.0)² = 8.33 × 10−3 kg
✓ 1
ALTERNATIVE: m ∝ 1/f², so m = 12.0 × (50.0/60.0)² = 8.33 g.
Mass removed = 12.0 − 8.33 = 3.7 g
✓ 1
Award [2] for CNA.
Part (d)(i)
2.1 mm
✓ 1
Accept 2.0–2.3 mm.
Part (d)(ii)
A small change in the pump's frequency causes a large fall in amplitude «roughly halving it for a 1 Hz change», and so a large fall in electrical output
✓ 1
Allow ECF from (d)(i).
Part (e)(i)
ω = 2π × 50.0 = 314 rad s−1
✓ 1
amax = ω²x0 = 314² × 4.0 × 10−3 = 395 m s−2
✓ 1
Fmax = ma = 0.0120 × 395 = 4.7 N
✓ 1
Award [3] for CNA.
Part (e)(ii)
The large force on the strip reverses direction twice in every cycle, so the strip is bent back and forth 50 times every second and the metal fatigues «destructive resonance»
✓ 1
OWTTE.
Part (f)
The driving force from the pump does work on the harvester, supplying energy on every cycle
✓ 1
The coil converts kinetic energy of the mass into electrical energy «and some energy is dissipated by friction and air resistance», removing energy on every cycle
✓ 1
The amplitude is constant because the energy supplied per cycle equals the energy removed per cycle
✓ 1
OWTTE.
Answers: (a) k = 1.18 × 103 N m−1 · (c) 3.7 g · (d)(i) 2.1 mm · (e)(i) 4.7 N (the remaining parts are explanations — see the table above)
Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; C.1 — the time period of a mass–spring system as given by T = 2π√(m/k); the defining equation of simple harmonic motion as given by a = −ω²x; A.2 — Newton's second law of motion Command term: Explain
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