A Level H1 Physics Waves Sound Light Quiz

A-Level Physics H1 Quiz - Waves Sound Light: Answer Key

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Section A: Short Answer Questions

1. [2 marks]

A transverse wave has oscillations perpendicular to the direction of energy transfer, whereas a longitudinal wave has oscillations parallel to the direction of energy transfer.

[B1] Correct identification of the oscillation direction for transverse waves (perpendicular to direction of propagation).
[B1] Correct identification of the oscillation direction for longitudinal waves (parallel to direction of propagation).

Teaching note: The key distinction is the orientation of particle displacement relative to the direction the wave travels. Electromagnetic waves (transverse) oscillate perpendicular to propagation; sound waves in air (longitudinal) oscillate along the direction of propagation.

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2. [1 mark]

The period of a wave is the time taken for one complete oscillation (or one complete cycle) of the wave.

[B1] Correct definition: time for one complete oscillation/cycle.

Teaching note: Period $T$ is measured in seconds and is related to frequency by $T = 1/f$.

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3. [2 marks]

Using $v = f\lambda$:

$v = 250 \times 1.36 = 340 \text{ m/s}$

[B1] Correct substitution into $v = f\lambda$.
[B1] Correct answer with unit: 340 m/s.

Teaching note: The wave equation $v = f\lambda$ relates speed, frequency, and wavelength. Always check that frequency is in Hz and wavelength is in metres to obtain speed in m/s.

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4. [2 marks]

When two or more waves meet at a point, the resultant displacement is equal to the vector sum of the individual displacements of each wave at that point.

[B1] Reference to two/more waves meeting/overlapping at a point.
[B1] Resultant displacement equals the sum (vector sum) of individual displacements.

Teaching note: Superposition is the fundamental principle underlying interference and stationary wave formation. It applies to all types of waves.

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5. [2 marks total]

(a) [1 mark] Amplitude is the maximum displacement from the equilibrium position — the arrow should extend vertically from the equilibrium line to the crest (or trough), labelled $A = 0.15$ m.

(b) [1 mark] Zero (or 0 m). Point P is at the equilibrium position, so its displacement at this instant is zero.

Teaching note: At the equilibrium position, displacement is zero but the particle has maximum velocity. Displacement is measured from the equilibrium (rest) position, not from the crest or trough.

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6. [2 marks]

Polarisation is the process by which the oscillations of a wave are restricted to a single plane (or direction). Only transverse waves can be polarised.

[B1] Correct description of polarisation (restriction of oscillations to one plane/direction).
[B1] Identification that only transverse waves can be polarised.

Teaching note: Longitudinal waves cannot be polarised because their oscillations are already along one direction (the direction of propagation). Polarisation is evidence of the transverse nature of electromagnetic waves.

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7. [2 marks]

Sound travels as a longitudinal wave through a medium by successive compressions and rarefactions. Steel is much stiffer (has a higher Young's modulus / elastic modulus) than air, so the restoring forces between particles are much greater. This allows the wave energy to be transmitted more rapidly, resulting in a higher wave speed.

[B1] Reference to the stiffness/elastic modulus of the medium being greater in steel.
[B1] Linking greater stiffness to faster transmission of disturbances/higher wave speed.

Teaching note: The speed of sound in a solid depends on $v = \sqrt{Y/\rho}$ where $Y$ is Young's modulus and $\rho$ is density. Although steel is denser than air, its Young's modulus is enormously larger, so the speed is much higher.

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8. [2 marks]

Two conditions for the formation of a stationary wave:

1. Two waves of the same frequency (and hence same wavelength) must travel in opposite directions along the same line.
2. The two waves must have approximately the same amplitude.

[B1] Same frequency/wavelength travelling in opposite directions.
[B1] Same (or similar) amplitude.

Teaching note: Stationary waves are formed by superposition of two identical waves travelling in opposite directions. This commonly occurs when a wave reflects back from a boundary and interferes with the incoming wave.

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9. [2 marks]

The general wave equation is $y = A \sin(\omega t - kx)$, where $\omega = 2\pi f$.

From the given equation: $\omega = 4.5 \times 10^3$ rad/s.

$f = \frac{\omega}{2\pi} = \frac{4.5 \times 10^3}{2\pi} = \frac{4500}{6.283} \approx 716 \text{ Hz}$

[B1] Correct identification that $\omega = 4.5 \times 10^3$ rad/s.
[B1] Correct calculation: 716 Hz (or 720 Hz to 2 s.f.).

Teaching note: In the wave equation $y = A \sin(\omega t - kx)$, the coefficient of $t$ is the angular frequency $\omega$, and the coefficient of $x$ is the wave number $k = 2\pi/\lambda$.

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10. [3 marks]

For a single-slit diffraction pattern, the condition for the first minimum is:

$a \sin\theta = \lambda$

where $a$ is the slit width and $\theta$ is the angle to the first minimum.

$a = \frac{\lambda}{\sin\theta} = \frac{6.0 \times 10^{-7}}{\sin 2.0°}$

$\sin 2.0° = 0.03490$

$a = \frac{6.0 \times 10^{-7}}{0.03490} = 1.72 \times 10^{-5} \text{ m}$

[B1] Correct formula: $a \sin\theta = \lambda$ (first minimum condition).
[B1] Correct substitution of values.
[B1] Correct answer: $1.7 \times 10^{-5}$ m (or 17 μm).

Teaching note: The single-slit diffraction formula for minima is $a \sin\theta = n\lambda$ where $n = 1, 2, 3, ...$ For the first minimum, $n = 1$. The small angle approximation is not needed here since $\sin\theta$ can be evaluated directly.

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Section B: Structured Questions

11. [5 marks total]

(a) [1 mark] Amplitude $A = 0.20$ m.
[B1] Read directly from the graph (maximum displacement from equilibrium).

(b) [1 mark] Wavelength $\lambda = 2.0$ m.
[B1] Read directly from the graph (distance between two consecutive crests or troughs).

(c) [2 marks]

$v = f\lambda \implies f = \frac{v}{\lambda} = \frac{8.0}{2.0} = 4.0 \text{ Hz}$

[B1] Correct substitution into $f = v/\lambda$.
[B1] Correct answer: 4.0 Hz.

(d) [1 mark] A node (N) should be marked at positions of zero displacement that remain zero at all times — these occur at $x = 0$, $x = 1.0$ m, $x = 2.0$ m, $x = 3.0$ m, and $x = 4.0$ m (every half-wavelength). Any one of these positions marked with N is acceptable.

[B1] Node correctly marked at a position corresponding to zero displacement in the stationary wave (at intervals of $\lambda/2 = 1.0$ m from either end).

Teaching note: In a stationary wave, nodes are points of zero displacement and antinodes are points of maximum displacement. Nodes occur every half-wavelength.

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12. [4 marks total]

(a) [2 marks]

$v = f\lambda \implies \lambda = \frac{v}{f} = \frac{340}{512} = 0.664 \text{ m}$

[B1] Correct substitution.
[B1] Correct answer: 0.66 m (or 0.664 m).

(b) [2 marks]

The frequency does not change when the wave moves from one medium to another — it is determined by the source. The wavelength increases because the speed increases ($v = f\lambda$, and $f$ is constant, so $\lambda$ must increase proportionally with $v$).

[B1] Frequency remains unchanged (determined by the source).
[B1] Wavelength increases because speed increases while frequency stays constant.

Teaching note: When a wave crosses a boundary between media, the frequency is always conserved because it is a property of the source. The speed and wavelength change according to the properties of the new medium.

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13. [4 marks total]

(a) [1 mark] Coherent sources are sources that have the same frequency and a constant phase difference (i.e., they maintain a fixed phase relationship).

[B1] Same frequency AND constant phase difference.

(b) [1 mark] At the first-order maximum, the path difference equals one wavelength: 0.50 m.

[B1] Path difference = $\lambda = 0.50$ m.

(c) [2 marks]

The intensity decreases for maxima further from the centre because the path difference increases, meaning the waves arriving at the detector from each source have travelled different distances. As the angle increases, the waves spread out more and the amplitude of each wave at the detector decreases. Additionally, for finite slit widths, the diffraction envelope modulates the intensity, reducing the intensity of higher-order maxima.

[B1] Reference to waves spreading out / amplitude decreasing with distance/angle.
[B1] Reference to the diffraction envelope effect or reduced amplitude at larger angles.

Teaching note: In a real double-slit experiment, each slit has finite width, so the interference pattern is modulated by the single-slit diffraction envelope. This causes the intensity of higher-order maxima to decrease.

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14. [6 marks total]

(a) [2 marks]

Third harmonic on a string fixed at both ends: 4 nodes (2 at the fixed ends + 2 intermediate) and 3 antinodes.

[B1] 4 nodes.
[B1] 3 antinodes.

Teaching note: For the $n$th harmonic on a string fixed at both ends, there are $(n+1)$ nodes and $n$ antinodes. The third harmonic ($n = 3$) has 4 nodes and 3 antinodes.

(b) [2 marks]

In the third harmonic, the string length contains 3 half-wavelengths:

$L = \frac{3\lambda}{2} \implies \lambda = \frac{2L}{3} = \frac{2 \times 1.20}{3} = 0.80 \text{ m}$

[B1] Correct relationship $L = 3\lambda/2$ (or equivalent).
[B1] Correct answer: 0.80 m.

(c) [2 marks]

$v = f\lambda \implies f = \frac{v}{\lambda} = \frac{240}{0.80} = 300 \text{ Hz}$

[B1] Correct substitution.
[B1] Correct answer: 300 Hz.

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15. [7 marks total]

(a) [2 marks]

400 lines per mm means:

$d = \frac{1}{400} \text{ mm} = \frac{1 \times 10^{-3}}{400} \text{ m} = 2.5 \times 10^{-6} \text{ m}$

[B1] Correct conversion (1/400 mm).
[B1] Correct answer: $2.5 \times 10^{-6}$ m (or 2.5 μm).

(b) [3 marks]

The diffraction grating equation is:

$d \sin\theta = n\lambda$

For the second-order maximum ($n = 2$):

$\sin\theta = \frac{n\lambda}{d} = \frac{2 \times 5.5 \times 10^{-7}}{2.5 \times 10^{-6}} = \frac{1.10 \times 10^{-6}}{2.5 \times 10^{-6}} = 0.440$

$\theta = \sin^{-1}(0.440) = 26.1°$

[B1] Correct formula $d\sin\theta = n\lambda$.
[B1] Correct substitution with $n = 2$.
[B1] Correct answer: 26° (or 26.1°).

(c) [2 marks]

The maximum possible order occurs when $\sin\theta = 1$:

$n_{\max} = \frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{5.5 \times 10^{-7}} = 4.55$

Since $n$ must be an integer, the highest observable order is $n = 4$.

[B1] Calculation of $n_{\max} = d/\lambda = 4.55$.
[B1] Correct conclusion: highest order is 4 (must round down since $n$ must be an integer and $\sin\theta$ cannot exceed 1).

Teaching note: The condition $\sin\theta \leq 1$ sets an upper limit on the diffraction order. Always round down to the nearest integer.

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16. [5 marks total]

(a) [2 marks]

Using Snell's law:

$n = \frac{\sin i}{\sin r} = \frac{\sin 50°}{\sin 31°} = \frac{0.7660}{0.5150} = 1.49$

[B1] Correct application of Snell's law.
[B1] Correct answer: 1.49 (or 1.5 to 2 s.f.).

(b) [2 marks]

$n = \frac{c}{v} \implies v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.49} = 2.01 \times 10^8 \text{ m/s}$

[B1] Correct formula $v = c/n$.
[B1] Correct answer: $2.0 \times 10^8$ m/s.

(c) [1 mark]

The emergent ray is parallel to the incident ray because the two refracting surfaces of the glass block are parallel. The refraction at the entry surface (bending towards the normal) is exactly reversed at the exit surface (bending away from the normal by the same angle), so the emergent ray is parallel to the incident ray but laterally displaced.

[B1] Correct explanation involving parallel surfaces and equal but opposite refraction at each surface.

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17. [6 marks total]

(a) [3 marks]

The fringe spacing formula for double-slit interference:

$\Delta y = \frac{\lambda D}{d}$

$\Delta y = \frac{650 \times 10^{-9} \times 1.5}{0.25 \times 10^{-3}} = \frac{9.75 \times 10^{-7}}{2.5 \times 10^{-4}} = 3.9 \times 10^{-3} \text{ m} = 3.9 \text{ mm}$

[B1] Correct formula $\Delta y = \lambda D/d$.
[B1] Correct substitution with all quantities in SI units.
[B1] Correct answer: 3.9 mm.

(b) [2 marks]

The fringe spacing decreases. Since $\Delta y = \lambda D/d$ and blue light has a shorter wavelength (450 nm < 650 nm), the fringe spacing is directly proportional to wavelength. A smaller wavelength means smaller fringe spacing.

[B1] Fringe spacing decreases.
[B1] Explanation linking shorter wavelength to smaller fringe spacing via $\Delta y \propto \lambda$.

(c) [1 mark]

Any one of:

• Increase the distance $D$ between the slits and the screen.
• Decrease the slit separation $d$.
• Use light of longer wavelength.

[B1] Any one valid change.

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Section C: Data Interpretation and Extended Response

18. [7 marks total]

(a) [2 marks]

The continuous spectrum (bremsstrahlung / braking radiation) is produced when high-speed electrons from the cathode are decelerated upon striking the anode target. As the electrons decelerate, they lose kinetic energy, which is emitted as X-ray photons. Because different electrons lose different amounts of energy (some lose all their energy in one collision, others lose it gradually), a continuous range of wavelengths is produced.

[B1] Reference to deceleration/braking of electrons at the anode.
[B1] Different electrons lose different amounts of energy, producing a continuous range of wavelengths.

(b) [3 marks]

The maximum energy of an X-ray photon equals the kinetic energy of the electron:

$eV = hf_{\max} = \frac{hc}{\lambda_{\min}}$

$\lambda_{\min} = \frac{hc}{eV} = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{1.6 \times 10^{-19} \times 1.24 \times 10^5}$

$\lambda_{\min} = \frac{1.989 \times 10^{-25}}{1.984 \times 10^{-14}} = 1.00 \times 10^{-11} \text{ m}$

[B1] Correct formula $\lambda_{\min} = hc/(eV)$.
[B1] Correct substitution.
[B1] Correct answer: $1.0 \times 10^{-11}$ m.

(c) [2 marks]

The sharp characteristic peak is produced when high-energy electrons from the cathode knock out inner-shell electrons from the anode atoms. Outer-shell electrons then fall into the inner-shell vacancies, emitting X-ray photons with specific energies (and hence specific wavelengths) that are characteristic of the anode material.

[B1] Reference to inner-shell electron being ejected from anode atoms.
[B1] Outer-shell electron fills the vacancy, emitting a photon of specific energy/wavelength characteristic of the target material.

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19. [7 marks total]

(a) [3 marks]

Using the Doppler effect formula for a source moving towards a stationary observer:

$f' = \frac{v}{v - v_s} \times f$

$f' = \frac{340}{340 - 25} \times 800 = \frac{340}{315} \times 800 = 1.0794 \times 800 = 863.5 \text{ Hz}$

[B1] Correct Doppler formula for source approaching observer.
[B1] Correct substitution.
[B1] Correct answer: 864 Hz (or 860 Hz to 2 s.f.).

(b) [2 marks]

For a source moving away from the observer:

$f'' = \frac{v}{v + v_s} \times f = \frac{340}{340 + 25} \times 800 = \frac{340}{365} \times 800 = 0.9315 \times 800 = 745.2 \text{ Hz}$

[B1] Correct formula for source receding.
[B1] Correct answer: 745 Hz (or 750 Hz to 2 s.f.).

(c) [2 marks]

As the car approaches, the sound waves are compressed (shorter wavelength, higher frequency) in front of the car. As the car recedes, the waves are stretched (longer wavelength, lower frequency) behind the car. At the instant the car passes the observer, the relative motion changes from approaching to receding, causing an abrupt switch from the higher observed frequency to the lower observed frequency.

[B1] Explanation of wave compression (higher frequency) when approaching.
[B1] Explanation of wave stretching (lower frequency) when receding, and the sudden change at the point of passing.

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20. [9 marks total]

(a) [1 mark]

$f^2$ for $T = 40$ N: $50.0^2 = 2500$ Hz² ✓ (already given)
$f^2$ for $T = 50$ N: $55.9^2 = 3124.81 \approx 3125$ Hz² ✓ (already given)

Both values are already completed in the table. [B1] Table is complete as shown.

(b) [3 marks]

The graph of $f^2$ against $T$ should be a straight line through the origin.

[B1] All points correctly plotted (within half a small square).
[B1] Appropriate scale chosen for both axes.
[B1] Best-fit straight line drawn through the origin.

Teaching note: The data shows that $f^2$ is directly proportional to $T$, which is consistent with the equation $f = \frac{1}{2L}\sqrt{T/\mu}$, giving $f^2 = T/(4L^2\mu)$.

(c) [2 marks]

Using two points on the best-fit line, e.g., $(0, 0)$ and $(50, 3125)$:

$\text{gradient} = \frac{\Delta f^2}{\Delta T} = \frac{3125 - 0}{50 - 0} = 62.5 \text{ Hz}^2/\text{N}$

[B1] Correct method (rise over run using two points on the line).
[B1] Correct answer: 62.5 Hz²/N (accept 62–63 depending on graph reading).

(d) [3 marks]

From the equation:

$f^2 = \frac{T}{4L^2\mu}$

So a graph of $f^2$ against $T$ has gradient $= \frac{1}{4L^2\mu}$.

$\text{gradient} = \frac{1}{4L^2\mu}$

$\mu = \frac{1}{4L^2 \times \text{gradient}} = \frac{1}{4 \times (0.80)^2 \times 62.5}$

$\mu = \frac{1}{4 \times 0.64 \times 62.5} = \frac{1}{160} = 6.25 \times 10^{-3} \text{ kg/m}$

[B1] Correct relationship: gradient $= 1/(4L^2\mu)$.
[B1] Correct substitution of $L = 0.80$ m and gradient value.
[B1] Correct answer: $6.25 \times 10^{-3}$ kg/m (or 6.25 g/m).

Teaching note: This question tests the ability to relate a theoretical equation to a linear graph and extract physical constants from the gradient. The key step is squaring the frequency equation to obtain a linear relationship.

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Mark Summary:

Section	Marks
A (Q1–10)	19
B (Q11–17)	24
C (Q18–20)	23
Total	50

Note: Individual question marks sum to 50. Section totals are approximate groupings.