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: wave speed on a string v = √(F_T/μ); v = fλ; intensity I = P/A; strings tied together have the same tension; polarizing filters are ideal, so unpolarized light leaves one filter with half its intensity. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A thick rope is tied to a thin rope under the same tension. An upright pulse is sent along the thick rope towards the knot. Which row describes the pulses after the pulse reaches the knot?
- (A) Reflected pulse upright; transmitted pulse upright
- (B) Reflected pulse inverted; transmitted pulse upright
- (C) Reflected pulse upright; transmitted pulse inverted
- (D) No reflected pulse; transmitted pulse upright
Answer and explanation
Answer: (A). At the same tension the thin rope has the smaller μ, so the wave travels faster in it. The pulse goes into a medium where its speed increases, so the reflected pulse is upright. The transmitted pulse is always upright.
- (B) is the result for a pulse going into a slower medium (thin rope to thick rope).
- (C) flips the transmitted pulse. The knot is lifted by the incident pulse and lifts the next rope, so the transmitted pulse cannot be inverted.
- (D) ignores reflection. Whenever the two media differ, part of the wave is reflected.
Question 2 (multiple choice · core)
A wave of frequency 12 Hz travels along a string at 30 m/s. It passes into a second string, where its speed is 18 m/s. What is its wavelength on the second string?
- (A) 1.5 m
- (B) 2.5 m
- (C) 4.2 m
- (D) 0.67 m
Answer and explanation
Answer: (A). The frequency stays at 12 Hz across the boundary, so λ₂ = v₂/f = 18 m/s ÷ 12 Hz = 1.5 m.
- (B) is the wavelength on the first string (30 ÷ 12). It assumes the wavelength stays fixed and the frequency changes.
- (C) uses the speed ratio upside down: 2.5 m × (30/18) = 4.17 m. A slower medium gives a shorter wavelength, not a longer one.
- (D) divides the frequency by the speed: 12 ÷ 18 = 0.67. The units of f/v are m⁻¹, not m.
Question 3 (multiple choice · foundation)
Which of these waves cannot be polarized?
- (A) Light from a desk lamp
- (B) A transverse wave on a stretched rubber cord
- (C) A sound wave in air
- (D) A radio wave from a transmitter
Answer and explanation
Answer: (C). Sound in air is a longitudinal wave. The air moves back and forth along the direction the wave travels, so there is no sideways direction for a filter or slit to select.
- (A) Light is transverse. Lamp light is unpolarized, but a filter can polarize it.
- (B) Any transverse wave on a cord can be made to oscillate in a single plane, for example by passing the cord through a narrow slot.
- (D) Radio waves are electromagnetic waves, which are transverse, so they can be polarized.
Question 4 (multiple choice · core)
An unpolarized light beam of power 0.24 W is spread evenly over an area of 3.0 × 10⁻⁴ m². It falls at right angles on one ideal polarizing filter. What is the intensity of the light that passes through?
- (A) 400 W/m²
- (B) 800 W/m²
- (C) 200 W/m²
- (D) 7.2 × 10⁻⁵ W/m²
Answer and explanation
Answer: (A). Incident intensity I₀ = P/A = 0.24 W ÷ 3.0 × 10⁻⁴ m² = 800 W/m². One ideal filter passes half of unpolarized light: I = 800 ÷ 2 = 400 W/m².
- (B) is the intensity before the filter. It forgets that the filter removes half of unpolarized light.
- (C) halves twice, as if there were two filters.
- (D) multiplies power by area instead of dividing: (0.24)(3.0 × 10⁻⁴) = 7.2 × 10⁻⁵. The units would be W·m², not W/m².
Question 5 (calculation · core)
String A (μ = 0.0050 kg/m) is tied to string B (μ = 0.0020 kg/m). The tension in both is 20 N. An upright pulse 0.30 m wide travels along string A towards the knot.
(a) Calculate the wave speed on each string. (b) State whether the reflected pulse is upright or inverted, with a reason. (c) Calculate the width of the transmitted pulse. (d) The free end of string A is now shaken at 50 Hz. Calculate the wavelength on each string.
Worked solution
- (a) v_A = √(20 N ÷ 0.0050 kg/m) = √4000 = 63.2 m/s. v_B = √(20 ÷ 0.0020) = √10 000 = 100 m/s.
- (b) The pulse goes from A into B, where the speed is greater. So the reflected pulse is upright.
- (c) The pulse takes 0.30 m ÷ 63.25 m/s = 4.74 × 10⁻³ s to pass the knot. The transmitted pulse leaves in the same time, so its width is (100 m/s)(4.74 × 10⁻³ s) = 0.474 m. Equivalently, width × (v_B/v_A) = 0.30 × 1.58 = 0.474 m.
- (d) f = 50 Hz on both strings. λ_A = 63.25 ÷ 50 = 1.26 m. λ_B = 100 ÷ 50 = 2.00 m.
Suggested mark points (4): 1 for both speeds; 1 for “upright” because the wave speeds up in B; 1 for a width of 0.474 m using the same time (or the speed ratio); 1 for both wavelengths using the same frequency on both strings.
Common error: giving the transmitted pulse the same width (0.30 m). The time to pass is fixed, not the width.
Question 6 (constructed response · core)
A student shines light from a lamp through two polarizing filters, one behind the other, onto a light meter. The light from the lamp is unpolarized, with intensity 50 W/m² at the first filter. The student keeps filter 1 fixed and slowly rotates filter 2 through 360°.
(a) Describe how the reading on the light meter changes as filter 2 turns through 360°. Give the largest and smallest intensities and say where they happen. (b) The student now replaces the lamp with a loudspeaker, the filters with a wooden board containing a narrow slot, and the light meter with a microphone. Turning the slot through 360° does not change the microphone reading. Explain what the two experiments together show about light and sound. (c) The student says: “Filter 1 cut the intensity to half because its axis happened to point the wrong way. If I turn filter 1, I can get all 50 W/m² through.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) After filter 1 the light is polarized along filter 1’s axis, with intensity 25 W/m². The reading is largest, 25 W/m², when the two axes are parallel (0° and 180°). It falls to zero when the axes are perpendicular (90° and 270°). So the reading goes through two maxima and two zeros in one full turn, changing smoothly between them.
(b) The light meter reading depends on the direction of the filter axis, so light can be polarized. Only transverse waves can be polarized, so light is transverse. The sound reading does not depend on the direction of the slot, so sound is not polarized by it. This fits sound being longitudinal: its disturbance is along the direction of travel, so there is no sideways direction to select.
(c) The claim is wrong. Unpolarized light has no preferred direction: its oscillation direction changes at random. Any axis direction passes the same share, half, so turning filter 1 gives 25 W/m² at every angle.
| Point | What earns it |
|---|---|
| 1 | Maximum of 25 W/m² when the axes are parallel |
| 1 | Zero when the axes are perpendicular, with two maxima and two minima in 360° |
| 1 | Light: reading depends on filter angle, so light is polarized, so it is transverse |
| 1 | Sound: no change with slot angle, consistent with a longitudinal wave that cannot be polarized |
| 1 | Rejects the claim because unpolarized light has no preferred direction, so any single filter passes half |
Do not award the (b) points for “light is transverse and sound is longitudinal” with no link to the observations.
Question 7 (constructed response · stretch)
Two strings are tied together under the same tension. An upright pulse 0.40 m wide travels along string 1 towards the knot. A short time later, a photograph shows an inverted pulse 0.40 m wide moving back along string 1 and an upright pulse 0.16 m wide moving along string 2.
(a) In which string is the wave speed greater? Give two separate pieces of evidence from the photograph. (b) Calculate the ratio v₂/v₁ and the ratio of the masses per unit length, μ₂/μ₁. (c) The free end of string 1 is now shaken at 15 Hz, and the wavelength on string 1 is 2.0 m. Calculate the wave speed on each string and the wavelength on string 2. (d) A student says: “The reflected pulse is as wide as the incident pulse, so it must carry the same energy.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) The speed is greater in string 1. Evidence 1: the reflected pulse is inverted, which happens when the wave goes into a slower medium. Evidence 2: the transmitted pulse is narrower (0.16 m against 0.40 m). Each pulse takes the same time to pass the knot, so the narrower pulse is the slower one.
(b) Widths are proportional to speeds: v₂/v₁ = 0.16 ÷ 0.40 = 0.40. With the same tension, v ∝ 1/√μ, so μ₂/μ₁ = (v₁/v₂)² = (1 ÷ 0.40)² = 6.25.
(c) v₁ = fλ₁ = (15 Hz)(2.0 m) = 30 m/s. v₂ = 0.40 × 30 = 12 m/s. The frequency is still 15 Hz, so λ₂ = 12 ÷ 15 = 0.80 m.
(d) The claim is wrong. The reflected pulse has the same width because it is still in string 1 and travels at the same speed. Width does not measure energy. The incident energy is shared between the reflected and transmitted pulses, so the reflected pulse carries less energy than the incident pulse and has a smaller amplitude.
| Point | What earns it |
|---|---|
| 1 | String 1 faster, with the inversion as evidence linked to “going into a slower medium” |
| 1 | Second piece of evidence: narrower transmitted pulse, linked to equal passing time |
| 1 | v₂/v₁ = 0.40 |
| 1 | μ₂/μ₁ = 6.25, using v ∝ 1/√μ at equal tension |
| 1 | v₁ = 30 m/s, v₂ = 12 m/s and λ₂ = 0.80 m, keeping f = 15 Hz |
| 1 | Rejects the claim: width is set by speed; energy is shared with the transmitted pulse, so the reflected pulse has less |
Carry forward an incorrect ratio from (b) into (c) once.
How did you do?
- Q1 or Q7(a) wrong: re-read “Inverted or upright? Follow the speed” and Figure 1 in the study guide.
- Q2, Q5 or Q7(c) wrong: revisit “Frequency stays the same; wavelength changes” and Worked example 1.
- Q3 or Q6 wrong: work through “Polarization” and Figure 2 again.
- Q4 wrong: revisit “Polarization and intensity” and Worked example 2.
Then tick off the topic checklist.