Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
Energy, damping and resonance
40 free, high-quality past-paper-style questions on Energy, damping and resonance for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 138 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U5-6-A-01Unit 5 · A2Easy1 mark
Which of the following is an example of a free oscillation?
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Notes
A free oscillation has no external periodic force acting once it has started, so the system vibrates at its own natural frequency. The plucked string is left to itself. In B, C and D an external periodic force keeps acting, so they are forced oscillations at the driving frequency.
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Answer A
Answer: A
Exam tip: Ask: is something still pushing it periodically? If yes, the oscillation is forced and follows the driver's frequency.
A mass on a spring of spring constant 40 N m−1 oscillates without damping with an amplitude of 0.050 m. What is the total energy of the oscillation?
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The total energy equals the elastic potential energy at the extreme: E = ½kA2 = ½ × 40 × 0.0502 = 0.050 J.
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Answer A
Answer: A
Common error: Forgetting to square the amplitude, or leaving out the ½. The energy stored in a spring is ½kx2, the area under its force–extension graph.
Air resistance reduces the amplitude of a pendulum's swing to 70% of its original value. What percentage of the original energy of the oscillation remains?
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Energy ∝ amplitude2, so the energy remaining is 0.702 = 0.49 of the original: 49%.
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Answer B
Answer: B
Exam tip: Whenever an amplitude changes, square the ratio to find what happens to the energy.
Which statement describes the condition for resonance of a forced oscillator?
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Resonance occurs when the driving frequency equals the natural frequency. Energy is then transferred from the driver most efficiently and the amplitude is a maximum.
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Answer B
Answer: B
Exam tip: Name both frequencies in a definition of resonance: the driving frequency and the natural frequency.
A glider of mass 0.20 kg on a horizontal air track is held between two springs. Together the springs act like a single spring of spring constant 32 N m−1. The glider is pulled 4.0 cm to one side and released. Friction is negligible.
Figure: glider held between two springs on an air track (not to scale)
(a)
Calculate the total energy of the oscillation.
(2)
(b)
Calculate the maximum speed of the glider.
(2)
(c)
State the position of the glider when its speed is a maximum.
(1)
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Part (a)
E = ½kA2 = ½ × 32 × 0.0402
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= 0.026 J (0.0256 J)
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Part (b)
½ × 0.20 × v2 = 0.0256
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v = 0.51 m s−1 (0.506 m s−1) (allow error carried forward)
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Part (c)
at the equilibrium (centre) position / where its displacement is zero
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Exam tip: All the energy is potential at the extreme and all kinetic at the centre. Equate ½kA2 to ½mvmax2, or use vmax = ωA as a check.
A trolley oscillates between two springs on a horizontal track. Damping is negligible. The graph shows how the potential energy of the system and the total energy vary with the displacement x of the trolley. What is the kinetic energy of the trolley when x = 2.0 cm?
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From the graph the total energy is 80 mJ and the potential energy at x = 2.0 cm is 20 mJ. With no damping the total is constant, so the kinetic energy is 80 − 20 = 60 mJ.
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Answer C
Answer: C
Exam tip: On an energy–displacement graph the kinetic energy at any point is the vertical gap between the total-energy line and the potential-energy curve.
A mass hangs from a spring. The top of the spring is attached to a vibration generator. The natural frequency of the mass–spring system is 4.0 Hz. The graph shows the displacement of the top of the spring (the driver) and the displacement of the mass once the motion has become steady. At what frequency is the mass oscillating?
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In a forced oscillation the mass vibrates at the driving frequency. From the graph both traces have a period of 0.40 s (four cycles in 1.6 s), so f = 1 ÷ 0.40 = 2.5 Hz. The natural frequency only affects how large the amplitude is.
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Answer B
Answer: B
Exam tip: Read a period over several cycles: count whole cycles between two points a long way apart and divide the time by the number of cycles.
A self-closing garden gate can swing both ways through its closed position. A damper on its hinge has three settings, P, Q and R. The graph shows how the angle of the gate from the closed position varies with time after it is released from 60°, for each setting. Which setting gives critical damping, and why?
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Critical damping returns a system to equilibrium in the shortest time without oscillating. Q never crosses the closed position and settles by about 1.2 s. P oscillates (light damping); R also avoids oscillating but takes much longer (heavy damping). So the answer is Q.
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Answer D
Answer: D
Exam tip: Critical and heavy damping both give no oscillation. Tell them apart by time: critical is the faster of the two.
Five pendulums hang from a horizontal string as shown. D has a heavy bob; P, Q, R and S have light bobs. The lengths are marked on the figure. D is pulled aside at right angles to the page and released. After a short time, which light pendulum is oscillating with the largest amplitude?
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D drives all the others at its own frequency through the string. The natural frequency of a pendulum depends on its length, so the one with the same length as D, S (0.60 m), has a natural frequency equal to the driving frequency. It resonates and has the largest amplitude.
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Answer D
Answer: D
Exam tip: In any coupled set-up, look for the object whose natural frequency matches the driver's: that is the one that resonates.
One end of a thin metal strip is clamped to a bench. The other end is attached to a vibration generator, whose frequency is changed in steps while the amplitude of the strip is measured. The graph shows the results.
(a)
State the natural frequency of the strip.
(1)
(b)
State the amplitude of the strip when the driving frequency is 14 Hz.
(1)
(c)
State the frequency at which the strip vibrates when the driving frequency is 14 Hz.
(1)
(d)
The vibration generator is switched off while the strip is vibrating. State the frequency at which the strip then vibrates.
(1)
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Part (a)
18 Hz (allow 17.5–18.5 Hz): the frequency at which the amplitude is greatest
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Part (b)
7.0 mm (allow 6.5–7.5 mm)
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Part (c)
14 Hz: a forced oscillation takes the frequency of the driver
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Part (d)
18 Hz / its natural frequency, because it is now a free oscillation
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Exam tip: Forced: the driver sets the frequency and the natural frequency sets the amplitude. Free: the system vibrates at its own natural frequency.
A pendulum bob of mass 0.25 kg hangs on a string of length 1.20 m. It is pulled aside to position A, as shown in the figure, and released. B is the lowest point of the swing. Air resistance is negligible.
(a)
Show that the height h of A above B is about 0.11 m.
(2)
(b)
Calculate the gravitational potential energy gained by the bob in moving from B to A.
(1)
(c)
Calculate the speed of the bob as it passes through B.
(2)
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Part (a)
h = L − L cos 25°
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= 1.20 − 1.20 × 0.906 = 0.112 m
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Part (b)
mgh = 0.25 × 9.81 × 0.112 = 0.28 J (0.276 J)
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Part (c)
½mv2 = 0.276 J, all the potential energy becomes kinetic energy
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v = √(2 × 0.276 ÷ 0.25) = 1.5 m s−1 (1.49 m s−1) (allow error carried forward)
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Exam tip: The height gained by a pendulum bob is not the string length: draw the vertical and use h = L(1 − cos θ).
The damping of a forced oscillator is increased, but nothing else is changed. Which row describes the effect on the peak of its resonance curve (amplitude against driving frequency)?
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More damping removes more energy per cycle, so the maximum amplitude is lower; the system responds over a wider range of frequencies, so the peak is broader; and the peak moves to a slightly lower frequency than the natural frequency.
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Answer D
Answer: D
Exam tip: Learn the three changes together: lower, broader, slightly lower frequency.
A 0.25 kg mass hangs from a spring of spring constant 64 N m−1. The top of the spring is moved up and down by a vibration generator whose frequency is slowly increased. At approximately what driving frequency is the amplitude of the mass greatest?
Figure: mass on a spring driven by a vibration generator (not to scale)
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Resonance occurs at the natural frequency. T = 2π√(m/k) = 2π√(0.25 ÷ 64) = 0.393 s, so f = 1/T = 2.5 Hz (2.55 Hz).
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Answer B
Answer: B
Common error: Giving √(k/m) as the frequency. That is the angular frequency ω in rad s−1; divide by 2π to get f in Hz.
A mass on a spring oscillates with simple harmonic motion and no damping. Its amplitude is A. What fraction of the total energy is kinetic energy when the displacement is A/2?
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Potential energy ∝ x2, so at x = A/2 it is (½)2 = ¼ of the total. The rest, 75%, is kinetic energy.
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Answer C
Answer: C
Exam tip: Potential energy = (x/A)2 × total energy. Kinetic energy is whatever is left.
The shock absorbers on a car are designed so that, after the car passes over a bump, the body returns to its normal height quickly without bouncing. Which type of damping is this closest to?
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Critical damping returns a system to equilibrium in the shortest time without oscillating. Light damping would leave the car bouncing; heavy damping would stop the bouncing but make the body return slowly and give a harsh ride over a series of bumps.
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Answer C
Answer: C
Exam tip: Critical damping: shortest time to equilibrium and no oscillation. Heavy damping avoids oscillation too, but takes longer.
A mass oscillates on a spring with simple harmonic motion and negligible damping. Its amplitude is 0.12 m and the total energy of the oscillation is 0.18 J.
(a)
Show that the spring constant is about 25 N m−1.
(2)
(b)
Calculate the potential energy and the kinetic energy of the oscillator when its displacement is 0.040 m.
(2)
(c)
Describe how the potential energy and the kinetic energy vary with displacement between −0.12 m and +0.12 m.
(2)
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Part (a)
0.18 = ½ × k × 0.122
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k = 0.36 ÷ 0.0144 = 25 N m−1
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Part (b)
potential energy = ½ × 25 × 0.0402 = 0.020 J
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kinetic energy = 0.18 − 0.020 = 0.16 J
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Part (c)
potential energy: a parabola / ∝ x2, zero at x = 0 and 0.18 J at ±0.12 m
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kinetic energy: an inverted parabola, 0.18 J at x = 0 and zero at ±0.12 m, so that the two always add to 0.18 J
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Model sketch
Figure: potential and kinetic energy against displacement
Exam tip: On an energy–displacement graph, the potential and kinetic curves are mirror images of each other in the horizontal line halfway up; at every displacement they add to the constant total.
A student films a mass oscillating on a spring inside a beaker of water and measures the amplitude at the start of successive cycles. Cycle number: 0, 1, 2, 3. Amplitude / mm: 50.0, 42.0, 35.3, 29.6.
Figure: amplitude at the start of each cycle
(a)
Show that the amplitude decreases by the same fraction in each cycle.
(2)
(b)
Calculate the percentage of the energy of the oscillation that is lost in each cycle.
(2)
(c)
Predict the amplitude at the start of cycle 6.
(1)
(d)
State what happens to the energy lost by the oscillator.
(1)
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Part (a)
42.0 ÷ 50.0 = 0.840 and 35.3 ÷ 42.0 = 0.840
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29.6 ÷ 35.3 = 0.839, so each ratio is about 0.84 / constant
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Part (b)
energy ∝ amplitude2, so energy ratio = 0.842 = 0.706
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percentage lost = 29% (29.4%)
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Part (c)
50.0 × 0.846 = 18 mm (17.6 mm)
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Part (d)
it is transferred to thermal energy of the water (and surroundings), by the drag force / viscous damping
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Common error: Taking the energy loss per cycle to be the same as the amplitude loss (16%). Energy depends on amplitude squared.
The drum of a washing machine is mounted on springs. The natural frequency of vertical vibration of the drum on its springs is 4.2 Hz. During a spin cycle the rotation rate of the drum increases steadily from zero to 1200 revolutions per minute. An uneven load of clothes makes the drum exert a periodic force on its mounting once per revolution.
(a)
Explain why the machine shakes violently for a short time while the drum speeds up.
(2)
(b)
Calculate the rotation rate, in revolutions per minute, at which this happens.
(1)
(c)
Explain why the vibration is small when the drum rotates at 1200 revolutions per minute, and why the machine is fitted with dampers.
(2)
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Part (a)
the uneven load provides a periodic driving force at the rotation frequency
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when the rotation frequency equals the natural frequency (4.2 Hz) resonance occurs, so the amplitude of vibration is a maximum
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Part (b)
4.2 × 60 = 250 rev min−1 (252)
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Part (c)
1200 rev min−1 = 20 Hz, far above the natural frequency, so the amplitude of the forced oscillation is small
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dampers transfer energy away / make the resonance peak lower (and broader), so the amplitude while passing through 4.2 Hz is limited
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Model sketch
Figure: amplitude of the drum against driving (rotation) frequency (not to scale)
Exam tip: A system driven far above its natural frequency barely moves. The danger is only while the driving frequency passes through the natural frequency.
A steel footbridge over a river has a deck of effective mass 6.0 × 104 kg. For vertical oscillations the deck behaves like a mass on a spring of spring constant 8.5 × 106 N m−1. People walking across take about two steps per second.
(a)
Calculate the natural frequency of vertical oscillation of the deck.
(2)
(b)
Explain why a large crowd walking across in step could make the deck oscillate with a large amplitude.
(2)
(c)
Dampers are fitted under the deck. State two effects this has on the response of the deck to driving at different frequencies.
(2)
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Part (a)
T = 2π√(6.0 × 104 ÷ 8.5 × 106) = 0.528 s
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f = 1/T = 1.9 Hz (1.89 Hz)
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Part (b)
the footsteps are a periodic driving force with a frequency (about 2 Hz) close to the natural frequency
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so resonance occurs: energy is transferred efficiently each cycle and the amplitude builds up to a large value
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Part (c)
the maximum amplitude (at resonance) is reduced
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the resonance peak becomes broader / the peak occurs at a slightly lower frequency
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Model sketch
Figure: effect of dampers on the response of the deck (not to scale)
Exam tip: The period formula T = 2π√(m/k) applies to any system that behaves like a mass on a spring, including large structures. Take care with the powers of ten.
A student uses a spring, a vibration generator and a signal generator to determine an unknown mass. For each known mass m hung on the spring she finds the resonant frequency f. Results: m = 0.100 kg, f = 2.63 Hz; 0.200 kg, 1.90 Hz; 0.300 kg, 1.57 Hz; 0.400 kg, 1.36 Hz; 0.500 kg, 1.22 Hz.
(a)
Describe how she should find the resonant frequency for one mass.
(2)
(b)
Calculate T2 for the 0.300 kg mass, where T = 1/f.
(1)
(c)
Her line of best fit on a graph of T2 against m passes through (0.100 kg, 0.145 s2) and (0.500 kg, 0.671 s2). Determine the spring constant.
(2)
(d)
With the unknown mass on the spring, the resonant frequency is 1.30 Hz. Use the line of best fit to determine the unknown mass.
(2)
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Answer
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Part (a)
increase the frequency of the signal generator slowly / in small steps while watching the amplitude of the mass (against a fixed scale)
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the resonant frequency is the frequency at which the amplitude is greatest; repeat approaching from higher frequencies and take a mean
m = 0.100 + (0.592 − 0.145) ÷ 1.315 = 0.44 kg (allow error carried forward)
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Model sketch
Figure: T² against m, with the unknown mass read from the line
Exam tip: Use the line of best fit, not the formula with k, to find the unknown mass: the line already allows for the mass of the spring, which gives it an intercept.
The graph shows how the velocity of a 0.50 kg mass oscillating on a spring varies with time. Damping is negligible. What is the total energy of the oscillation?
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The greatest speed, 0.40 m s−1, occurs as the mass passes through the equilibrium position, where all the energy is kinetic. E = ½mvmax2 = ½ × 0.50 × 0.402 = 0.040 J.
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Answer B
Answer: B
Exam tip: At the equilibrium position the potential energy is zero, so the total energy of an undamped oscillator equals its maximum kinetic energy.
An oscillator is driven by a periodic force of constant amplitude. The graph shows how its amplitude varies with the driving frequency for three different amounts of damping, X, Y and Z. Which curve is for the least damping, and what is the approximate natural frequency of the oscillator?
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Less damping gives a higher, sharper resonance peak, so X has the least damping. With light damping the peak is at (very nearly) the natural frequency: about 5.0 Hz. More damping lowers and broadens the peak and moves it to a slightly lower frequency, as Z shows.
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Answer A
Answer: A
Exam tip: Take the natural frequency from the most lightly damped curve: its peak lies closest to the natural frequency.
The graphs show how the force F varies with extension x when three materials P, Q and R are loaded and then unloaded. An engineer needs a material for braces that will reduce the amplitude of a building's oscillations during an earthquake. Which material is most suitable, and why?
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R is ductile. The area under its loading curve is much larger than the area under its unloading line, so most of the work done on it is not returned: it is dissipated (as thermal energy). Removing energy from the oscillating building each cycle reduces its amplitude. Q returns all its energy, so it removes none; P is brittle and fractures.
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Answer A
Answer: A
Exam tip: Energy dissipated = area under the loading curve − area under the unloading curve. A large difference means good damping.
A damped pendulum is released with an amplitude of 8.0 cm. The graph shows how the energy of its oscillation decreases with time. What is the amplitude of the oscillation at t = 6.0 s?
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From the graph the energy falls from 64 mJ to 16 mJ, to ¼ of its starting value. Energy ∝ amplitude2, so the amplitude falls by a factor √4 = 2: 8.0 ÷ 2 = 4.0 cm.
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Answer C
Answer: C
Common error: Scaling the amplitude by the same factor as the energy. Take the square root of the energy ratio to get the amplitude ratio.
A 0.80 kg block oscillates horizontally on the end of a spring on a smooth surface. Damping is negligible. The graph shows how the kinetic energy of the block varies with its displacement x.
(a)
Use the graph to show that the spring constant is 50 N m−1.
(2)
(b)
Calculate the frequency of the oscillation.
(2)
(c)
Determine the displacements at which the kinetic energy equals the potential energy.
(2)
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Part (a)
total energy = maximum kinetic energy = 90 mJ and amplitude = 6.0 cm, both read from the graph
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k = 2E/A2 = 2 × 0.090 ÷ 0.0602 = 50 N m−1
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Part (b)
f = (1/2π)√(k/m) = (1/2π)√(50 ÷ 0.80)
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= 1.3 Hz (1.26 Hz)
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Part (c)
each is then half the total, so ½kx2 = 45 mJ / KE = 45 mJ on the graph
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x = ±A/√2 = ±4.2 cm (allow ±4.1 to ±4.3 cm)
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Exam tip: On a kinetic-energy graph the potential energy at any displacement is the gap between the curve and the total energy (the peak value).
A mass on a spring of spring constant 18 N m−1 oscillates vertically with a small card attached to it, so that air resistance damps the motion. The graph shows how the displacement of the mass varies with time.
(a)
Calculate the energy of the oscillation at t = 0.
(1)
(b)
Use the graph to calculate the energy transferred from the oscillation during the first complete cycle.
(2)
(c)
Calculate the mean rate at which energy is transferred from the oscillation during this cycle.
(1)
(d)
The amplitude falls by the same fraction in every cycle. Determine the number of complete cycles after which the energy of the oscillation is first less than 10% of its value at t = 0.
(2)
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Answer
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Part (a)
½kA2 = ½ × 18 × 0.1002 = 0.090 J
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Part (b)
amplitude after one cycle = 8.0 cm, so energy = ½ × 18 × 0.0802 = 0.0576 J
period = 0.80 s from the graph; 0.0324 ÷ 0.80 = 0.041 W (0.0405 W) (allow error carried forward)
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Part (d)
energy is multiplied by 0.802 = 0.64 each cycle
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0.645 = 0.107 and 0.646 = 0.069, so after 6 cycles
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Common error: Taking the energy to fall by 20% per cycle because the amplitude does. Energy depends on amplitude squared: 0.802 = 0.64, a 36% fall each cycle.
In an experiment to determine an unknown mass, a student hangs known masses m on a light spring attached to a vibration generator and finds the resonant frequency f for each mass. He plots a graph of f2 against 1/m, shown in the figure with his line of best fit.
(a)
Show that the graph is expected to be a straight line through the origin.
(2)
(b)
Use the graph to determine the spring constant.
(2)
(c)
With the unknown mass on the spring, the resonant frequency is 1.60 Hz. Determine the unknown mass.
(2)
(d)
Suggest one way the student could make the resonant frequency more precise.
(1)
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Answer
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Part (a)
resonant frequency = natural frequency, f = (1/2π)√(k/m)
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so f2 = (k/4π2) × (1/m), which has the form y = mx with a constant gradient k/4π2
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Part (b)
gradient = 7.60 ÷ 10.0 = 0.760 (Hz2 kg)
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k = 4π2 × 0.760 = 30 N m−1 (30.0 N m−1)
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Part (c)
f2 = 2.56 Hz2, so 1/m = 2.56 ÷ 0.760 = 3.37 kg−1 (allow reading from the graph, 3.3–3.4)
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m = 0.30 kg (0.297 kg) (allow error carried forward)
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Part (d)
use smaller frequency steps near the peak / approach the peak from above and below and average / use a fixed scale or video to judge the largest amplitude
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Exam tip: Rearrange the equation into y = mx + c with the plotted quantities as y and x: the gradient then tells you which constant you can find.
A model building of mass 1.5 kg is fixed to a shaking table, which moves from side to side with a constant amplitude at a frequency that can be varied. The graph shows how the amplitude of the top of the model varies with the shaking frequency, without and with a damper fitted.
(a)
State the natural frequency of the model.
(1)
(b)
The model behaves like a 1.5 kg mass on a spring. Calculate the effective spring constant.
(2)
(c)
Describe two ways in which the curve with the damper differs from the curve without it.
(2)
(d)
Explain why the damper reduces the amplitude at resonance.
(1)
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Answer
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Part (a)
2.0 Hz, the peak of the curve without the damper
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Part (b)
k = 4π2f2m = 4π2 × 2.02 × 1.5
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= 240 N m−1 (237 N m−1)
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Part (c)
the peak amplitude is lower (5.0 mm instead of 25 mm)
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the peak is broader / at a slightly lower frequency
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Part (d)
the damper transfers energy from the oscillation each cycle (as thermal energy), so the amplitude builds up to a smaller value at which the energy supplied by the driver per cycle equals the energy dissipated
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Exam tip: When asked to compare resonance curves, give the three features: height of the peak, width of the peak and position of the peak.
The graph shows the vertical displacement of one corner of a car body after the car passes over a bump, for two different shock absorbers, P and Q. That corner of the car can be modelled as a mass of 320 kg on a spring.
(a)
State the type of damping shown by each curve, giving a reason for each.
(2)
(b)
Use curve P to determine the spring constant of the suspension at this corner. Assume the period of P equals the natural period.
(2)
(c)
With shock absorber P, the car is driven along a road with ridges across it every 15 m. Calculate the speed at which the car body would oscillate with the largest amplitude, and explain why.
(2)
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Answer
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Part (a)
P: light damping / under-damped, because the body oscillates about equilibrium with a decreasing amplitude
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Q: critical damping, because the body returns to equilibrium quickly (in less than about one period) without oscillating
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Part (b)
period = 0.80 s from the graph
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k = 4π2m/T2 = 4π2 × 320 ÷ 0.802 = 2.0 × 104 N m−1
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Part (c)
the ridges give a periodic driving force; the amplitude is greatest at resonance, when the time between ridges equals the natural period, 0.80 s
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v = 15 ÷ 0.80 = 19 m s−1 (18.75 m s−1)
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Exam tip: A regularly spaced pattern on a road becomes a driving frequency when you move over it: f = speed ÷ spacing.
In an experiment to find an unknown mass, a student measures the resonant frequency of a spring loaded with known masses m and plots a graph of T2 against m, where T = 1/f. The gradient of the line is 1.5 s2 kg−1. What is the spring constant of the spring?
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Answer
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T = 2π√(m/k) gives T2 = (4π2/k)m, so the gradient is 4π2/k. k = 4π2 ÷ 1.5 = 26 N m−1 (26.3 N m−1).
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Answer C
Answer: C
Exam tip: Write the equation in the form of the graph first: T2 = (4π2/k) × m. Then the gradient is whatever multiplies the x-variable.
An undamped pendulum has a period of 0.50 s. At what frequency does its kinetic energy vary?
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Answer
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The oscillation frequency is 1/0.50 = 2.0 Hz. The kinetic energy is a maximum each time the bob passes through the centre — twice per cycle — so it varies at twice the oscillation frequency: 4.0 Hz.
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Answer D
Answer: D
Exam tip: Kinetic and potential energy never go negative: each reaches a maximum twice per cycle, so energy–time graphs have double the frequency of the displacement–time graph.
In a resonance experiment a mass of (0.250 ± 0.001) kg is hung on a spring. The resonant frequency is found to be (1.62 ± 0.02) Hz. The mass of the spring may be neglected.
(a)
Calculate the percentage uncertainty in the resonant frequency.
(1)
(b)
Calculate the percentage uncertainty in T2, where T is the period.
(1)
(c)
Calculate the spring constant and its absolute uncertainty.
(3)
(d)
Suggest why the resonant frequency is harder to find precisely if the mass is surrounded by a card that increases air resistance.
(1)
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Answer
Mark
Notes
Part (a)
0.02 ÷ 1.62 × 100 = 1.2%
✓ 1
Part (b)
T2 = 1/f2, so 2 × 1.2% = 2.5% (2.47%)
✓ 1
Part (c)
k = 4π2mf2 = 4π2 × 0.250 × 1.622 = 25.9 N m−1
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percentage uncertainty in k = 0.4% + 2.5% = 2.9%
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k = (25.9 ± 0.7) N m−1
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Part (d)
more damping makes the resonance peak lower and broader, so the amplitude changes little near the peak and the frequency of maximum amplitude is harder to judge
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Exam tip: When a quantity is raised to a power, multiply its percentage uncertainty by that power; then add the percentage uncertainties of quantities that are multiplied or divided.
Tall buildings in earthquake zones are designed so that the oscillations of the building during an earthquake do not become large enough to cause serious damage. Some designs include steel braces made of a ductile steel that is intended to deform plastically during strong shaking.
(a)
Explain how the design of a building can reduce the amplitude of its oscillations during an earthquake. Your answer should refer to resonance, damping and the plastic deformation of ductile materials.
(6)
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Answer
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Notes
Indicative content:
—
the ground motion acts as a periodic driving force on the building, which has its own natural frequency
✓ 1
if the driving frequency is close to the natural frequency, resonance occurs and the amplitude becomes very large
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so the building is designed with a natural frequency away from the main earthquake frequencies, e.g. by using flexible base isolators
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dampers transfer energy from the oscillating building, e.g. to thermal energy, so the amplitude at resonance is reduced / the resonance peak is lower and broader
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when ductile steel deforms plastically, work is done that is not recovered, so energy is dissipated each cycle (as thermal energy)
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removing energy reduces the amplitude because energy ∝ amplitude2; a ductile material can do this repeatedly without fracturing, unlike a brittle one
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Exam tip: In an extended answer, link each design feature to the physics: avoiding resonance changes the frequency match; damping and plastic deformation remove energy, which lowers the amplitude.
One corner of a car can be modelled as a mass of 350 kg supported by a spring of spring constant 2.8 × 104 N m−1, with a shock absorber connected alongside the spring. After driving over a bump, the corner of the car is displaced 0.060 m from its equilibrium position and is momentarily at rest.
(a)
Calculate the natural frequency of this mass–spring system.
(2)
(b)
Calculate the energy of the oscillation at this moment.
(1)
(c)
The shock absorber provides close to critical damping. Describe the motion of the car body after the bump, and state what happens to the 50 J of energy.
(2)
(d)
When the shock absorber is badly worn, the amplitude decreases by 40% in each cycle. Calculate the energy of the oscillation after two complete cycles.
(2)
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Answer
Mark
Notes
Part (a)
T = 2π√(350 ÷ 2.8 × 104) = 0.702 s
✓ 1
f = 1/T = 1.4 Hz (1.42 Hz)
✓ 1
Part (b)
½ × 2.8 × 104 × 0.0602 = 50 J (50.4 J)
✓ 1
Part (c)
it returns to its equilibrium position in the shortest time without oscillating
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the energy is transferred to thermal energy of the oil / shock absorber and surroundings
✓ 1
Part (d)
amplitude after two cycles = 0.060 × 0.602 = 0.0216 m / energy ratio per cycle = 0.602 = 0.36
A student hangs known masses m on a spring, finds the resonant frequency f for each using a vibration generator, and plots a graph of T2 against m, where T = 1/f. The graph shows her results and her line of best fit. With an unknown mass on the spring, the resonant frequency is 1.25 Hz. What is the unknown mass?
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Answer
Mark
Notes
Gradient = (0.88 − 0.24) ÷ (0.50 − 0.10) = 1.60 s2 kg−1; intercept = 0.24 − 1.60 × 0.10 = 0.080 s2. For the unknown mass T2 = 1 ÷ 1.252 = 0.64 s2, so m = (0.64 − 0.080) ÷ 1.60 = 0.35 kg. Reading the line directly at T2 = 0.64 s2 gives the same mass.
✓ 1
Answer A
Answer: A
Exam tip: Use the line itself, intercept included, to convert a measured T2 into a mass. The intercept comes from the mass of the spring, which also oscillates.
A 0.20 kg mass hangs from a spring and is driven by a vibration generator. The graph shows how the amplitude of the mass varies with the driving frequency. Damping is light. A further 0.25 kg is added to the mass. At approximately what driving frequency is the amplitude now greatest?
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Answer
Mark
Notes
From the graph resonance is at 3.0 Hz, which is (very nearly) the natural frequency. f = (1/2π)√(k/m), so f ∝ 1/√m. The new mass is 0.45 kg, so f = 3.0 × √(0.20 ÷ 0.45) = 3.0 × ⅔ = 2.0 Hz.
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Answer B
Answer: B
Exam tip: Use a ratio: the spring constant cancels, so you only need how f depends on m. A larger mass always lowers the resonant frequency.
A mass on a spring oscillates horizontally on a smooth surface with an amplitude of 8.0 cm. Damping is negligible. The graph shows how the elastic potential energy stored in the spring varies with time.
(a)
Explain why the period of the oscillation is 0.50 s.
(2)
(b)
Calculate the spring constant.
(2)
(c)
Calculate the mass.
(2)
(d)
Calculate the maximum speed of the mass.
(1)
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Answer
Mark
Notes
Part (a)
the potential energy is a maximum at each extreme of the motion / at both +A and −A, which happens twice per cycle
✓ 1
the maxima are 0.25 s apart, so one cycle takes 2 × 0.25 = 0.50 s
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Part (b)
total energy = 0.32 J = ½kA2
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k = 2 × 0.32 ÷ 0.0802 = 100 N m−1
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Part (c)
T = 2π√(m/k), so m = kT2/4π2
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= 100 × 0.502 ÷ 4π2 = 0.63 kg (0.633 kg)
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Part (d)
½ × 0.633 × v2 = 0.32, so v = 1.0 m s−1 (1.01 m s−1) (or v = 2πA/T) (allow error carried forward)
✓ 1
Common error: Reading the time between energy maxima as the period. Energy reaches a maximum twice in every cycle, so the energy graph repeats at twice the frequency of the motion.
Steel braces in a building are designed to deform plastically when the building sways during an earthquake. The graph shows how the force in one brace varies with its extension during one complete cycle of the building's oscillation.
(a)
Explain why the area enclosed by the loop is the energy dissipated in the brace during one cycle.
(1)
(b)
Determine the energy dissipated in the brace during one cycle.
(2)
(c)
After the ground stops shaking, the building keeps swaying. At the start of one cycle the energy of its oscillation is 30 kJ. The building has 6 such braces, and during this cycle each goes through approximately the loop shown. Assume the braces are the only cause of energy loss. Calculate the amplitude of the building's oscillation at the end of this cycle as a percentage of its amplitude at the start of the cycle.
(2)
(d)
Suggest why the braces are made of a ductile material rather than a brittle one.
(1)
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Answer
Mark
Notes
Part (a)
the area under the curve while extending is the work done on the brace and the area under the curve while it returns is the work it gives back; the difference, the enclosed area, is not recovered and is dissipated as thermal energy
✓ 1
Part (b)
the loop is a parallelogram of base 8.0 mm and height 120 kN
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energy = 0.0080 × 1.20 × 105 = 960 J (allow 940–980 J by counting squares)
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Part (c)
energy dissipated = 6 × 960 = 5760 J, so energy left = 30 000 − 5760 = 24 240 J
a ductile material can deform plastically through large extensions many times, absorbing energy, whereas a brittle material would fracture with almost no plastic deformation / energy dissipated
✓ 1
Common error: Reducing the amplitude by the same percentage as the energy. Amplitude is proportional to √energy, so losing about 19% of the energy reduces the amplitude by only about 10%.
A student uses a spring, a vibration generator and a signal generator to find an unknown mass. With the unknown mass on the spring she measures the amplitude of oscillation at a series of driving frequencies. The graph shows her results and a smooth curve through them. The mass of the spring is negligible.
(a)
Use the graph to state the resonant frequency.
(1)
(b)
Earlier, the student found that the spring extends by 12.3 cm when a 0.300 kg mass hangs from it at rest. Show that the spring constant is about 24 N m−1.
(1)
(c)
Calculate the unknown mass.
(2)
(d)
Explain why the student took readings at closer intervals of frequency between 1.30 Hz and 1.50 Hz.
(1)
(e)
The student attaches a large card to the mass, which increases the damping, and repeats the experiment. Explain how this would affect the graph and the value obtained for the unknown mass.
(2)
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Answer
Mark
Notes
Part (a)
1.40 Hz (allow 1.39–1.41 Hz)
✓ 1
Part (b)
k = mg/x = 0.300 × 9.81 ÷ 0.123 = 23.9 N m−1
✓ 1
Part (c)
m = k/(4π2f2) = 23.9 ÷ (4π2 × 1.402)
✓ 1
= 0.31 kg (0.309 kg) (allow error carried forward)
✓ 1
Part (d)
the amplitude changes rapidly with frequency near the peak, so closely spaced readings are needed to locate the peak / resonant frequency precisely
✓ 1
Part (e)
the peak would be lower and broader, so the resonant frequency would be harder to locate precisely
✓ 1
the peak would move to a slightly lower frequency, so the calculated mass would be slightly too large (the card also adds a little mass)
✓ 1
Exam tip: In a resonance practical, take more readings where the graph changes fastest: near the peak. Light damping gives a sharp peak and a precise resonant frequency.
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