These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: c = 3.00 × 10⁸ m/s; on a string v = √(F_T/μ) with μ = m/ℓ; strings are uniform, and the tension is the same all along the string. All data are invented for practice. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A small buoy floats on a lake. A series of water waves passes it, with the crests moving east. Which statement best describes the buoy and the waves?
- (A) The buoy is carried east with the crests, at the speed of the waves.
- (B) The buoy moves mainly up and down near one place, while energy moves east.
- (C) The buoy does not move at all, because water waves carry no energy.
- (D) The buoy drifts steadily east, because each wave moves a layer of water east.
Answer and explanation
Answer: (B). A wave transfers energy without transferring matter. The water under the buoy, and the buoy with it, oscillates around one position as each crest and trough passes. The disturbance, and its energy, moves east.
- (A) treats the buoy as if it were part of the travelling pattern. The pattern moves; the medium does not travel with it.
- (C) is wrong twice: the buoy does move (it bobs), and the energy to make it bob is carried by the waves.
- (D) describes a current, not a wave. In the wave model used in this course, there is no net flow of the medium.
Question 2 (multiple choice · foundation)
A long spring lies on a table along a north–south line. A student repeatedly pushes and pulls the south end of the spring along the north–south direction. Which row describes the wave produced?
- (A) Transverse; each coil moves east and west.
- (B) Longitudinal; each coil moves north and south around its own equilibrium position.
- (C) Longitudinal; each coil travels to the north end of the spring.
- (D) Transverse; each coil moves north and south.
Answer and explanation
Answer: (B). The disturbance (the push and pull) is north–south, and the wave travels north along the spring. The two directions are parallel, so the wave is longitudinal. Each coil oscillates around its own position; regions of bunched and spread coils travel north.
- (A) describes a sideways shake, which would make a transverse wave. That is not what the student does.
- (C) has the right type but says the coils travel with the wave. Only the compressions and the energy travel.
- (D) mixes up the labels: north–south motion along a north–south spring is parallel to the travel, so it cannot be transverse.
Question 3 (multiple choice · core)
A pulse travels along a string at speed v. The string is replaced by one with half the mass per length, and the tension is doubled. What is the new pulse speed?
- (A) 4v
- (B) 2v
- (C) v
- (D) √2 v
Answer and explanation
Answer: (B). v = √(F_T/μ). The ratio F_T/μ becomes (2F_T)/(μ/2) = 4F_T/μ, four times larger. The speed is the square root, so it becomes √4 = 2 times larger: 2v.
- (A) forgets the square root and multiplies v by the factor 4 directly.
- (C) treats the two changes as cancelling. Halving μ increases the speed, just as doubling F_T does; they work in the same direction here.
- (D) includes only one of the two changes (for example, only the doubled tension).
Question 4 (multiple choice · core)
Four students make claims about sound and light. Which claim is correct?
- (A) Sound travels faster through air at 30 °C than through the same air at 0 °C.
- (B) Sound and light both travel at 3.00 × 10⁸ m/s in a vacuum.
- (C) A loud sound travels through air faster than a quiet sound with the same frequency.
- (D) Light from a lamp inside a glass jar cannot be seen once all the air is pumped out of the jar.
Answer and explanation
Answer: (A). In a given medium, the speed of sound increases with the temperature of the medium.
- (B) is wrong because sound is a mechanical wave and cannot travel in a vacuum at all. Only EM waves travel at c in a vacuum.
- (C) confuses amplitude with speed. A louder sound has a larger amplitude and carries more energy, but its speed is set by the medium.
- (D) treats light as a mechanical wave. Light is electromagnetic and needs no medium, so the lamp stays visible. (A ringing bell in the jar would become hard to hear.)
Question 5 (calculation · core)
A guitar-style string is 1.5 m long and has a mass of 12 g. It is stretched to a tension of 50 N.
(a) Calculate the speed of a pulse on the string. (b) Calculate the time a pulse takes to travel from one end of the string to the other. (c) The tension is changed so that a pulse crosses the string in half the time. Find the new tension. (d) A student sends a pulse that is twice as tall. State the effect, if any, on your answer to (b).
Worked solution
- μ = m/ℓ = 0.012 kg ÷ 1.5 m = 8.0 × 10⁻³ kg/m.
- (a) v = √(F_T/μ) = √(50 N ÷ 0.0080 kg/m) = √6250 m²/s² = 79.1 m/s.
- (b) t = ℓ/v = 1.5 m ÷ 79.06 m/s = 0.0190 s (19.0 ms).
- (c) Half the time means twice the speed. Since v ∝ √F_T, the tension must be 2² = 4 times larger: F_T = 4 × 50 N = 200 N. (Check: √(200 ÷ 0.0080) = 158 m/s, and 1.5 ÷ 158 = 0.0095 s.)
- (d) No effect. The speed depends on F_T and μ only, so the travel time is still 0.0190 s. The taller pulse carries more energy.
Suggested mark points (4): 1 for μ = 8.0 × 10⁻³ kg/m with mass in kg; 1 for v = 79.1 m/s and t = 0.0190 s; 1 for 200 N with reasoning from v ∝ √F_T (or by finding the new v and using F_T = μv²); 1 for “no effect” with the reason that speed depends on the medium, not the amplitude.
Common error: using the total mass, √(50 ÷ 0.012) = 64.5 m/s. The formula needs mass per length. Another error is answering (c) with 100 N, which forgets the square root.
Question 6 (constructed response · core)
A loudspeaker at one end of a long tube sends a steady sound wave along the tube. The normal air pressure in the tube is 101 300 Pa. At one instant, sensors along the tube record these pressure changes from normal:
| Position x (m) | 0 | 0.25 | 0.50 | 0.75 | 1.00 |
|---|---|---|---|---|---|
| Pressure change (Pa) | 0 | +0.8 | 0 | −0.8 | 0 |
(a) State the position, between 0 and 1.00 m, of a compression and of a rarefaction. Justify your answer. (b) State the pressure amplitude of the wave and the largest pressure (in Pa) at any point in the tube. (c) Sketch a graph of pressure change against position for 0 ≤ x ≤ 1.00 m at this instant. (d) The volume is turned up, at the same frequency. Compare the new wave with the original in terms of amplitude, energy carried and speed. Justify each comparison.
Worked solution and suggested Marlbridge rubric
(a) A compression is a region of higher than normal pressure: x = 0.25 m (+0.8 Pa). A rarefaction is a region of lower than normal pressure: x = 0.75 m (−0.8 Pa). At the compression the air layers are crowded together; at the rarefaction they are spread apart.
(b) Amplitude = largest change from the equilibrium pressure = 0.8 Pa. Largest pressure = 101 300 Pa + 0.8 Pa = 101 300.8 Pa.
(c) A smooth, wave-shaped curve: zero at x = 0, a maximum of +0.8 Pa at 0.25 m, zero at 0.50 m, a minimum of −0.8 Pa at 0.75 m and zero again at 1.00 m. Axes labelled with quantities and units; the horizontal axis is position, not time.
(d) Louder means a larger amplitude (the pressure swings further from 101 300 Pa). A larger amplitude means the wave carries more energy. The speed is unchanged, because it depends on the air in the tube (and its temperature), not on the amplitude.
| Point | What earns it |
|---|---|
| 1 | Compression at 0.25 m and rarefaction at 0.75 m, linked to higher and lower than normal pressure |
| 1 | Amplitude 0.8 Pa (not 1.6 Pa) and largest pressure 101 300.8 Pa |
| 1 | Sketch: smooth curve through the five points with maximum at 0.25 m and minimum at 0.75 m, labelled axes |
| 1 | Larger amplitude and more energy, linked to the sound being louder |
| 1 | Same speed, because speed depends on the medium and not on the amplitude |
Do not award the second point for 1.6 Pa: that is the change from the lowest to the highest pressure, which is twice the amplitude.
Question 7 (constructed response · stretch)
A student hangs different masses from a string that passes over a pulley, to set the tension F_T. For each tension, she measures the speed v of a pulse along the horizontal part of the string.
| F_T (N) | 2.0 | 4.0 | 6.0 | 8.0 |
|---|---|---|---|---|
| v (m/s) | 20.2 | 28.1 | 34.8 | 39.9 |
(a) Explain why a graph of v² against F_T is expected to be a straight line through the origin. State what its gradient represents. (b) Calculate v² for each tension. Use the data to find the mass per length μ of the string. (c) Another student claims: “The data show that doubling the tension doubles the pulse speed.” Evaluate this claim using the data. (d) Predict the pulse speed when the tension is 12 N.
Worked solution and suggested Marlbridge rubric
(a) Squaring v = √(F_T/μ) gives v² = (1/μ)F_T. This has the form y = mx with y = v² and x = F_T, so the graph is a straight line through the origin with gradient 1/μ.
(b) v² values (m²/s²): 408, 790, 1211, 1592. A best-fit line gives a gradient of about 199 m²/s² per N, so μ = 1/199 ≈ 5.0 × 10⁻³ kg/m. (A line through the origin gives 1/199.8; a two-point gradient from the first and last points gives 197.3. All give μ = 5.0 × 10⁻³ kg/m to 2 significant figures.)
(c) The claim is not supported. From 4.0 N to 8.0 N, the tension doubles but the speed rises from 28.1 m/s to 39.9 m/s, a factor of 1.42. From 2.0 N to 4.0 N the factor is 1.39. Both are close to √2 ≈ 1.41, not 2. This matches v ∝ √F_T.
(d) v = √(F_T/μ) = √(12 N ÷ 0.0050 kg/m) = √2400 ≈ 49 m/s. (Using the gradient directly: v² = 199 × 12 ≈ 2390, v ≈ 49 m/s.)
| Point | What earns it |
|---|---|
| 1 | Rearranges to v² = F_T/μ and identifies the gradient as 1/μ |
| 1 | Correct v² values (at least three correct) |
| 1 | Gradient from a line of best fit (or a sound two-point method using well-separated points) about 197–200 |
| 1 | μ ≈ 5.0 × 10⁻³ kg/m, with units, from μ = 1/gradient |
| 1 | Rejects the claim with numbers: doubling F_T multiplies v by about 1.4 (≈ √2), not 2 |
| 1 | Prediction of about 49 m/s (accept 48–50 m/s) |
Accept a method for (d) that reads the gradient from the graph. Carry forward an incorrect μ from (b) into (d) once.
How did you do?
- Q1 or Q2 wrong: re-read “Energy moves, matter stays” and “Transverse and longitudinal waves” in the study guide.
- Q3, Q5 or Q7 wrong: revisit “Pulses on a string” and Worked example 1. Check mass per length and the square root.
- Q4 wrong: compare mechanical and electromagnetic waves, and re-read “Sound and temperature”.
- Q6 wrong: look again at Figure 2 and the section on amplitude.
Then tick off the topic checklist.