Sound — Practice · Grade 12 Science

Chapter: Waves, Sound & Optics

What this practice covers

By the end of Sound you should be able to:

10 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: longitudinal wave, pitch, loudness, medium, Doppler effect, resonance, standing wave.

Problem 1

Using the speed of sound as 340 m/s, how far away is a storm if thunder is heard 3 seconds after the lightning flash?
Answer and explanation

1020 m

distance = speed × time = 340 × 3 = 1,020 m.

Problem 2

The pitch of a sound is determined by its…
  1. amplitude
  2. loudness
  3. the medium's color
  4. frequency
Answer and explanation

frequency

Pitch depends on frequency — higher frequency means higher pitch.

Problem 3

As an ambulance moves away from you, the siren's pitch drops because…
  1. the waves stretch out behind it, lowering the frequency you hear
  2. the sound doubles in speed
  3. the pitch rises as the waves compress
  4. the pitch stays exactly the same
Answer and explanation

the waves stretch out behind it, lowering the frequency you hear

For a receding source, the wavefronts stretch out, so the observed wavelength grows and the frequency (pitch) drops — the Doppler effect.

Problem 4

A sound wave in air has a wavelength of 0.68 m. Using the speed of sound as 340 m/s, find its frequency.
Answer and explanation

500 Hz

f = v ÷ λ = 340 ÷ 0.68 = 500 Hz.

Problem 5

Explain what resonance is, and give one everyday example of it.
Answer and explanation

Resonance is the large build-up of vibration that happens when a system is driven at its natural frequency. Pushing a swing in time with its natural back-and-forth rhythm makes it go higher and higher; other examples include a singer shattering a wine glass by matching its natural frequency, or a guitar string producing a loud clear note.

Problem 6

A train sounds an 800 Hz whistle while moving away from you at 60 m/s. Using the speed of sound as 340 m/s, what frequency do you hear? (f′ = f × v ÷ (v + vₛ))
Answer and explanation

680 Hz

The source recedes, so f′ = f × v ÷ (v + vₛ) = 800 × 340 ÷ (340 + 60) = 800 × 340 ÷ 400 = 800 × 0.85 = 680 Hz.

Problem 7

An echo returns in 0.20 s with sound speed 340 m/s. Find wall distance. Show the reasoning, not just the conclusion.
Hints

Beat frequency is distinct from the approximately 442 Hz pitch region.

Answer and explanation

340×0.20/2=34 m, including the round trip.

Problem 8

Why does sound speed change between materials?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Elastic and inertial properties of the medium affect propagation; speed is not fixed by loudness alone.

Problem 9

A decibel difference represents a logarithmic intensity ratioA decibel difference represents a logarithmic intensity ratioLevel difference 10 log₁₀(I/I₀) (dB)010203002505007501000Intensity ratio I/I₀ (linear axis)10× intensity → +10 dB;100× → +20 dB.
Calculated selected intensity-ratio points; not measured sound exposure or a safety threshold.
Interpret the supplied visual. What intensity ratio corresponds to a 30 dB increase?
Answer and explanation

10^(30/10)=1000, provided the levels use the same intensity reference.

A sound-intensity level is β=10log₁₀(I/I₀) for a stated reference intensity I₀. Differences in decibels compare ratios: a 20 dB increase means 100 times the intensity, not 20 times. Decibels are not a direct linear loudness scale. The plotted points show selected calculated ratios on a linear horizontal axis; between-point connections are a guide rather than an exact smooth logarithmic curve.

10^(30/10)=1000, provided the levels use the same intensity reference.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

Two sound intensity levels differ by 15 dB. Find their intensity ratio. A separate echo returns in 0.30 s at 340 m/s; find reflector distance.
Answer and explanation

The intensity ratio is 10^1.5≈31.6. Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.

The intensity ratio is 10^1.5≈31.6.

Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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