Secondary 3 Physics Waves Sound Light Quiz

Secondary 3 Physics Quiz - Waves Sound Light: Answer Key

Total Marks: 40 marks

Section A: Multiple Choice

1. B — water wave travelling across the surface of a pond [1 mark]

• Water waves are transverse: particles vibrate perpendicular to wave direction.
• Sound waves (A, C, D) are longitudinal: particles vibrate parallel to wave direction.
• Common mistake: Confusing all waves with being transverse. Remember sound needs a medium and particles compress/rarefy in the direction of travel.

2. C — $6.7 \times 10^{-7} \text{ m}$ [1 mark]

• Using $v = f\lambda$, so $\lambda = \frac{v}{f} = \frac{3.0 \times 10^8}{4.5 \times 10^{14}} = 6.667 \times 10^{-7} \text{ m} \approx 6.7 \times 10^{-7} \text{ m}$
• Note: This is red light (visible range ~400–700 nm). The answer is approximately 667 nm.

3. C — frequency [1 mark]

• Pitch corresponds to frequency: higher frequency = higher pitch.
• Amplitude (A) determines loudness, not pitch.
• Key concept: Frequency is the number of complete vibrations per second, measured in hertz (Hz).

4. C — frequency [1 mark]

• When light enters a denser medium (air to glass): speed decreases, wavelength decreases ($v = f\lambda$ with $f$ constant), direction changes (refraction).
• Frequency stays constant because it is determined by the source, not the medium.
• Underlying principle: The wave equation $v = f\lambda$; if $v$ changes and $\lambda$ changes proportionally, $f$ remains unchanged.

5. B — they spread out in circular arcs [1 mark]

• This is diffraction: waves spread out after passing through a gap comparable to their wavelength.
• Condition for noticeable diffraction: Gap size ≈ wavelength or smaller. Straight wavefronts become circular when the gap is small.

Section B: Short Answer and Structured Questions

6. [2 marks]

(a) Amplitude: The maximum displacement of a particle from its equilibrium position. [1 mark]

• For transverse waves: maximum height of crest or depth of trough from rest position.
• For longitudinal waves: maximum compression or rarefaction density change.

(b) Period: The time taken for one complete wave cycle (or oscillation) to pass a point. [1 mark]

• Related to frequency by $T = \frac{1}{f}$. Unit: seconds (s).

7. [2 marks]

Using $v = f\lambda$:

$v = 250 \times 1.32 = 330 \text{ m/s}$ [1 mark for formula, 1 mark for answer with unit]

• Step-by-step:
  1. Identify: frequency $f = 250 \text{ Hz}$, wavelength $\lambda = 1.32 \text{ m}$
  2. Apply wave equation: $v = f\lambda$
  3. Substitute: $v = 250 \times 1.32 = 330$
  4. State unit: $\text{m/s}$

This is the typical speed of sound in air at room temperature.

8. [3 marks]

(a) Amplitude = 3.0 cm (or $0.03 \text{ m}$) [1 mark]

• Read directly from diagram: vertical distance from equilibrium to crest.

(b) Frequency calculation: [2 marks]

First convert amplitude-related values to metres: amplitude given as 3.0 cm, but we need wavelength: $\lambda = 8.0 \text{ cm} = 0.08 \text{ m}$

Using $v = f\lambda$:

$f = \frac{v}{\lambda} = \frac{0.48}{0.08} = 6.0 \text{ Hz}$ [1 mark for conversion and substitution, 1 mark for final answer]

• Critical step: Convert cm to m before calculation. Wavelength = 8.0 cm = 0.08 m.
• Speed was given as 0.48 m/s, so consistent units are essential.

9. [2 marks]

Sound waves require a medium to travel because they are longitudinal mechanical waves. [1 mark]

Sound travels by particles vibrating and transferring energy through collisions. [0.5 mark]

In a vacuum there are no particles, so there is no mechanism for the vibrations to be transmitted. [0.5 mark]

• Key distinction: Electromagnetic waves can travel through vacuum; mechanical waves (including sound) cannot.
• Exam tip: Always mention "medium" and "particles" in explanations about sound requiring matter.

10. [4 marks]

(a) Speed of sound: [2 marks]

For an echo, sound travels to the wall AND back: total distance = $2 \times 85 = 170 \text{ m}$

$v = \frac{d}{t} = \frac{170}{0.50} = 340 \text{ m/s}$ [1 mark for doubling distance, 1 mark for calculation]

• Common error: Using 85 m instead of 170 m gives 170 m/s — wrong because the sound makes a round trip.

(b) Two distinct sounds: [2 marks]

As the student walks towards the wall, the time for the echo to return decreases. [1 mark]

When very close to the wall, the direct sound and echo arrive with very short time separation, becoming distinguishable or merging; while farther apart, the time gap between direct sound and echo is longer. [1 mark]

Alternatively accepted: Standing at 85 m gives clear echo; moving closer reduces echo time delay. At some positions the student may hear the original clap (from hands) and the reflected echo as separate sounds if time gap > 0.1 s (persistence of hearing).

11. [2 marks]

(a) Law of reflection: The angle of incidence equals the angle of reflection ($i = r$). [1 mark]

• Both angles are measured between the ray and the normal, not between ray and mirror surface.

(b) Angle $r = 35°$ and reflected ray drawn symmetrically about normal. [1 mark]

• Using $i = r = 35°$. The reflected ray should be drawn in the upper right quadrant, making 35° with normal ON.
• Marking note: Ray must have correct direction (away from mirror), correct angle, and arrowhead.

12. [3 marks]

(a) Common property: (Any one of) [1 mark]

• All travel at the same speed in a vacuum ($3.0 \times 10^8 \text{ m/s}$)
• All are transverse waves
• All can travel through a vacuum
• All transfer energy without transferring matter
• All obey the wave equation $v = f\lambda$

(b) Uses: [2 marks]

13. [4 marks]

(a) Ray diagram: [2 marks]

Two standard rays from top of object:

• Ray 1: Parallel to principal axis → refracts through focal point F on far side [1 mark]
• Ray 2: Through centre of lens C → continues undeviated [0.5 mark]

Image formed where backward extensions of diverging rays meet (virtual image on same side as object). [0.5 mark for locating image correctly]

Image pending generation: diagram for Q13.

(b) Image characteristics: (Any two of) [2 marks]

• Virtual (cannot be projected on screen)
• Upright (same orientation as object)
• Magnified (larger than object)
• Located on same side of lens as object

14. [3 marks]

(a) Why wavelength changes: [2 marks]

The wave slows down in shallow water because the water depth restricts particle motion. [1 mark]

Since frequency stays constant (determined by source), and $v = f\lambda$, a decrease in $v$ causes a proportional decrease in $\lambda$. [1 mark]

• Key reasoning chain: depth decrease → speed decrease → wavelength decrease (frequency unchanged).

(b) Frequency stays the same. [1 mark]

• Frequency is determined by the wave source (e.g., vibration frequency of dipper in ripple tank), not by the medium.

15. [4 marks]

(a) P-wave time: [2 marks]

$t_P = \frac{d}{v_P} = \frac{120}{6.0} = 20 \text{ s}$ [1 mark for formula, 1 mark for answer with unit]

• Straightforward substitution. Distance in km, speed in km/s, so time in s.

(b) Time difference: [2 marks]

First find S-wave time: $t_S = \frac{120}{4.0} = 30 \text{ s}$ [1 mark]

Time difference: $\Delta t = t_S - t_P = 30 - 20 = 10 \text{ s}$ [1 mark]

• Physical significance: P-waves arrive first. Seismologists use this time difference to locate earthquake epicentres.

Section C: Extended Response

16. [3 marks]

Experiment to measure speed of sound using echo method:

Apparatus: [1 mark]

• Stopwatch or time interval recorder
• Measuring tape or trundle wheel
• Large vertical wall or building (flat, hard surface) with clear space in front
• Wooden blocks or两块木板 (to clap together, or use starting pistol/electronic sound source)

Measurements: [1 mark]

• Measure the perpendicular distance $d$ from student to wall using measuring tape (e.g., 50–100 m for clear time measurement)
• Produce a sharp sound (clap blocks) and start timer simultaneously
• Stop timer when echo is heard
• Repeat several times and calculate average time $t$

Calculation: [1 mark]

• Total distance travelled by sound = $2d$ (to wall and back)
• Speed of sound: $v = \frac{2d}{t}$

Improvements/accuracy notes:

• Distance should be large enough ($>50$ m) for time measurement >0.3 s
• Use two observers with synchronized stopwatches, or electronic timing
• Take multiple readings and average
• Avoid background noise and wind

17. [6 marks]

(a) $v = f\lambda$ (wave speed = frequency × wavelength) [1 mark]

(b) Wavelength in air: [2 marks]

$\lambda = \frac{v}{f} = \frac{340}{440} = 0.773 \text{ m}$ (or 77.3 cm, or ~0.77 m to 2 s.f.) [1 mark for formula, 1 mark for calculation]

(c) New wavelength and explanation: [3 marks]

$\lambda_{water} = \frac{v_{water}}{f} = \frac{1480}{440} = 3.36 \text{ m}$ [1 mark for calculation]

Explanation why frequency is constant: [2 marks]

• Frequency is determined by the source of the wave (the vibrating object producing the sound). [1 mark]

• When sound enters a different medium, the speed changes because the new medium has different elastic properties and density, but the source vibration rate stays the same. [0.5 mark]

• Since $v = f\lambda$ and $f$ is fixed, the wavelength $\lambda$ must change proportionally with $v$ to maintain the equality. [0.5 mark]

• Numerical check: Water speed (1480) is about 4.35× air speed (340), so wavelength increases by same factor: 0.773 × 4.35 ≈ 3.36 m. ✓

18. [4 marks]

(a) Why white light separates into colours: [2 marks]

White light consists of many different frequencies (colours), each with a different wavelength. [1 mark]

The refractive index of glass depends on wavelength (dispersion): shorter wavelengths (violet) are slowed more than longer wavelengths (red), causing different angles of refraction for each colour. [1 mark]

• Key term: "Dispersion" — the separation of white light into its component colours.

(b) Why violet is deviated more: [2 marks]

Violet light has a shorter wavelength and higher frequency than red light. [1 mark]

The refractive index of glass for violet ($n = 1.53$) is greater than for red ($n = 1.51$). Using Snell's law, a higher refractive index means light bends more toward the normal at each refraction, resulting in greater overall deviation through the prism. [1 mark]

• Calculation check: For incident angle 40°, using $\frac{\sin i}{\sin r} = n$:
  • Red: $\sin r = \frac{\sin 40°}{1.51} = 0.426$, so $r = 25.2°$
  • Violet: $\sin r = \frac{\sin 40°}{1.53} = 0.420$, so $r = 24.9°$
  • Violet bends more at first surface; similar at second surface, cumulative effect gives greater deviation.

19. [4 marks]

(a) Longitudinal nature: [2 marks]

In this wave, the particles of the spring vibrate parallel to the direction of wave travel. [1 mark]

The diagram shows compressions (C) where coils are close together and rarefactions (R) where coils are spread out — the displacement arrows point along the spring's length, not perpendicular to it. [1 mark]

• Contrast with transverse: In a transverse wave, particle displacement would be perpendicular to wave direction (e.g., string waved up and down).

(b) Wave speed: [2 marks]

$v = f\lambda = 2.5 \times 0.80 = 2.0 \text{ m/s}$ [1 mark for formula, 1 mark for answer with unit]

20. [4 marks]

(a) Intensity is inversely proportional to the square of the distance: [1 mark]

$I \propto \frac{1}{d^2}$ or $I = \frac{k}{d^2}$ where $k$ is constant, or "$I \times d^2$ = constant"

(b) Completed table and analysis: [3 marks]

[1 mark for correct $d^2$ column, 1 mark for correct $I \times d^2$ column]

Analysis: [1 mark]

The product $I \times d^2$ is approximately constant (≈ 80 W in each case), confirming that $I \propto \frac{1}{d^2}$. This is the inverse square law for intensity from a point source.

• Small variation (80.1) due to rounding in given data; students should note consistency.
• Physical basis: Sound energy spreads over spherical area $A = 4\pi d^2$, so intensity (power/area) decreases as $1/d^2$.

Total: 40 marks