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: T = 1/f; v = fλ; at one location the displacement is A cos(2πft) = A cos(2πt/T); at one instant it is A cos(2πx/λ); on a string v = √(F_T/μ); waves are sinusoidal with constant amplitude; use radians in cosine equations. All data are invented for practice. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A small loudspeaker cone completes 2400 vibrations in 4.0 s. Which row gives the frequency and the period of the sound it produces?
- (A) 600 Hz and 1.7 × 10⁻³ s
- (B) 2400 Hz and 4.2 × 10⁻⁴ s
- (C) 600 Hz and 600 s
- (D) 1.7 × 10⁻³ Hz and 600 s
Answer and explanation
Answer: (A). f = 2400 ÷ 4.0 s = 600 Hz, and T = 1/f = 1 ÷ 600 Hz = 1.67 × 10⁻³ s (equivalently 4.0 s ÷ 2400).
- (B) ignores the 4.0 s and treats the number of vibrations as the frequency.
- (C) gets f right but then writes T = f instead of T = 1/f.
- (D) swaps the two: it calls 1/600 the frequency and 600 the period.
Question 2 (multiple choice · foundation)
A snapshot of a wave on a rope shows a crest 5.0 cm above the undisturbed rope and the nearest trough 5.0 cm below it. The horizontal distance from that crest to the trough is 0.30 m. What are the wavelength and the amplitude?
- (A) λ = 0.60 m, A = 5.0 cm
- (B) λ = 0.30 m, A = 10 cm
- (C) λ = 0.60 m, A = 10 cm
- (D) λ = 0.30 m, A = 5.0 cm
Answer and explanation
Answer: (A). Crest to the nearest trough is half a wavelength, so λ = 2 × 0.30 m = 0.60 m. The amplitude is measured from equilibrium to the crest: 5.0 cm.
- (B) makes both common errors: crest to trough taken as λ, and the full 10 cm height taken as A.
- (C) has λ right but uses the crest-to-trough height (2A) as the amplitude.
- (D) has A right but takes crest to trough as a full wavelength.
Question 3 (multiple choice · core)
A vibrator sends a periodic wave along a string at fixed tension. The frequency of the vibrator is doubled; the amplitude stays the same. What happens to the wavelength and the speed of the wave?
- (A) The wavelength halves; the speed stays the same.
- (B) The wavelength doubles; the speed stays the same.
- (C) The wavelength stays the same; the speed doubles.
- (D) The wavelength halves; the speed halves.
Answer and explanation
Answer: (A). The speed on a string depends only on F_T and μ, which have not changed. So v is the same, and λ = v/f halves when f doubles.
- (B) inverts the relationship: λ is inversely proportional to f at fixed v.
- (C) assumes the source controls the speed. The medium sets the speed; the source sets f.
- (D) changes both, which would break v = fλ: (2f)(λ/2) = fλ, so v cannot halve if λ halves and f doubles.
Question 4 (multiple choice · core)
Two sound waves travel through the same air. Wave X has a frequency of 400 Hz. Wave Y has a frequency of 800 Hz and the same amplitude. Which statement is correct?
- (A) Y has the higher pitch and half the wavelength of X, and both travel at the same speed.
- (B) Y is louder than X because it has the higher frequency, and both have the same wavelength.
- (C) Y travels twice as fast as X, so it has the same wavelength.
- (D) Y has twice the period of X and a lower pitch.
Answer and explanation
Answer: (A). Pitch follows frequency, so Y sounds higher. In the same air both travel at the same speed, so λ = v/f gives Y half the wavelength. (Y also carries energy at a greater rate, because it has the higher frequency at the same amplitude.)
- (B) links loudness to frequency. Loudness depends on amplitude, and the amplitudes are equal. The wavelengths differ.
- (C) assumes speed depends on frequency. In one medium it does not.
- (D) gets the period backwards: T = 1/f, so Y has half the period of X, and a higher pitch.
Question 5 (calculation · core)
At one point on a string, the displacement is y = (0.040 m) cos(2πt / 0.0125 s). The wavelength of the wave is 1.5 m.
(a) State the amplitude and the period, and calculate the frequency. (b) Calculate the speed of the wave. (c) Calculate the displacement of this point at t = 0.0050 s. (d) Write an equation for the displacement along the string, as a function of x, at the instant when the point at x = 0 has its maximum displacement. Use it to find the displacement at x = 0.50 m at that instant.
Worked solution
- (a) Compare with y = A cos(2πt/T): A = 0.040 m and T = 0.0125 s. f = 1/T = 80 Hz.
- (b) v = fλ = 80 Hz × 1.5 m = 120 m/s.
- (c) 2πt/T = 2π × (0.0050 ÷ 0.0125) = 2π × 0.40 = 0.80π rad. y = 0.040 × cos(0.80π) = 0.040 × (−0.809) = −0.0324 m (−3.24 cm).
- (d) At that instant the point at x = 0 is at +A, so y = (0.040 m) cos(2πx / 1.5 m). At x = 0.50 m: 2π × 0.50/1.5 = 2π/3 rad, cos(2π/3) = −0.50, so y = −0.020 m.
Suggested mark points (4): 1 for A, T and f = 80 Hz; 1 for v = 120 m/s; 1 for −0.0324 m with the calculator in radians; 1 for the position equation with λ = 1.5 m and y = −0.020 m.
Common error: evaluating cos(0.80π) in degree mode, which gives a value very close to 1 and y ≈ +0.040 m. A displacement near +A at 0.40 of a period cannot be right: the point should be below equilibrium, close to its trough.
Question 6 (constructed response · core)
A wave travels along a long string at 24 m/s. The source vibrates at 60 Hz with an amplitude of 2.0 cm. At t = 0 the point at x = 0 is at its maximum upward displacement.
(a) Calculate the wavelength and the period. (b) Sketch, with numbered scales on both axes: (i) displacement against position at t = 0, for 0 ≤ x ≤ 1.2 m; (ii) displacement against time at x = 0, for 0 ≤ t ≤ 0.050 s. (c) The frequency of the source is tripled, with the same tension and amplitude. State the new wavelength and period, and describe how each of your sketches would change. (d) Compare the energy carried by the wave before and after the change in (c).
Worked solution and suggested Marlbridge rubric
(a) λ = v/f = 24 m/s ÷ 60 Hz = 0.40 m. T = 1/f = 1 ÷ 60 Hz = 0.0167 s (16.7 ms).
(b) (i) A cosine curve starting at +2.0 cm at x = 0, with crests at 0, 0.40, 0.80 and 1.2 m: exactly 3 wavelengths across 1.2 m. Troughs of −2.0 cm at 0.20, 0.60 and 1.0 m. (ii) A cosine curve starting at +2.0 cm at t = 0, with crests every 0.0167 s: exactly 3 periods across 0.050 s. Both vertical scales run from −2.0 to +2.0 cm.
(c) The speed is unchanged (same F_T and μ). New f = 180 Hz, so λ = 24 ÷ 180 = 0.133 m and T = 1/180 = 5.56 × 10⁻³ s. Both sketches keep the same height (amplitude 2.0 cm) but now show 9 cycles instead of 3: the curves are squeezed horizontally by a factor of 3.
(d) The amplitude is the same but the frequency is higher, so the wave carries more energy (each point of the string goes through the same oscillation three times as often).
| Point | What earns it |
|---|---|
| 1 | λ = 0.40 m and T = 0.0167 s |
| 1 | Sketch (i): starts at +A, amplitude 2.0 cm, three full cycles over 1.2 m, axis labelled position in m |
| 1 | Sketch (ii): starts at +A, amplitude 2.0 cm, three full cycles over 0.050 s, axis labelled time in s |
| 1 | New λ = 0.133 m and T = 5.56 ms, with the speed stated as unchanged and the reason |
| 1 | Sketches: same amplitude, three times as many cycles on each |
| 1 | More energy, linked to the higher frequency at the same amplitude |
Do not award a sketch point if its horizontal axis is labelled with the wrong quantity (for example, λ marked on a time axis).
Question 7 (constructed response · stretch)
A string has mass per length μ = 2.0 × 10⁻³ kg/m. It is held at a tension of 7.2 N and driven by a vibrator at 150 Hz.
(a) Calculate the wave speed and the wavelength. (b) Starting from v = √(F_T/μ) and v = fλ, derive an expression for λ in terms of F_T, μ and f. (c) A student wants waves of wavelength 0.80 m on this string. Describe two different single changes that would achieve this, and calculate the value needed in each case. (d) Another student says: “Increasing the tension increases the frequency of the waves on the string.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) v = √(7.2 N ÷ 0.0020 kg/m) = √3600 = 60 m/s. λ = v/f = 60 ÷ 150 = 0.40 m.
(b) Set the two expressions for v equal: fλ = √(F_T/μ), so λ = (1/f)√(F_T/μ). (Check: (1/150) × 60 = 0.40 m.)
(c) The wavelength must double.
- Change 1: lower the frequency. At the same tension, v = 60 m/s, so f = v/λ = 60 ÷ 0.80 = 75 Hz (half the original).
- Change 2: raise the tension. At 150 Hz, v must be fλ = 150 × 0.80 = 120 m/s, twice as fast. Since v ∝ √F_T, F_T must be 4 times larger: F_T = μv² = 0.0020 × 120² = 28.8 N.
(d) The claim is incorrect. The frequency of every point on the string is set by the vibrator, so it stays 150 Hz. Increasing the tension increases the wave speed, and since f is fixed, λ = v/f gets longer. The expression from (b) shows this: F_T appears with λ, and f does not depend on F_T.
| Point | What earns it |
|---|---|
| 1 | v = 60 m/s and λ = 0.40 m |
| 1 | Correct derivation of λ = (1/f)√(F_T/μ), starting from the two given equations |
| 1 | Change 1: frequency reduced to 75 Hz at the same tension |
| 1 | Change 2: tension raised to 28.8 N (four times) at the same frequency, with working |
| 1 | States that the source sets the frequency, so f is unchanged |
| 1 | Explains that higher tension gives a higher speed and therefore a longer wavelength |
Accept “use a string with one quarter of the mass per length (5.0 × 10⁻⁴ kg/m)” as a valid alternative for one of the changes in (c).
How did you do?
- Q1 or Q2 wrong: re-read “Waves that repeat” in the study guide, checking T = 1/f and where λ and A are measured.
- Q3, Q4 or Q7 wrong: revisit “Speed, frequency and wavelength”: the medium sets v, the source sets f.
- Q5 wrong: work through “Equations for a sinusoidal wave” and Worked example 1, in radians.
- Q6 wrong: compare the two graphs in Figure 1 and their horizontal axes.
Then tick off the topic checklist.