Sound Waves — Practice · Grade 8 Science

Chapter: Waves, Sound & Light

What this practice covers

By the end of Sound Waves you should be able to:

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

Vocabulary used: longitudinal wave, compression, rarefaction, pitch, medium, model, evidence, system boundary, assumption.

Problem 1

Using the speed of sound as 340 m/s, how far away is a thunderstorm if thunder is heard 4 seconds after the lightning flash?
Answer and explanation

1360 m

distance = speed × time = 340 × 4 = 1,360 m.

Problem 2

Sound is what type of wave?
  1. transverse
  2. longitudinal
  3. electromagnetic
  4. standing only
Answer and explanation

longitudinal

Sound particles vibrate in the same direction the wave travels, making it a longitudinal wave.

Problem 3

Increasing the frequency of a sound wave makes the pitch…
  1. higher
  2. lower
  3. louder only
  4. quieter only
Answer and explanation

higher

Higher frequency corresponds to higher pitch.

Problem 4

Sound CANNOT travel through…
  1. water
  2. steel
  3. a vacuum (empty space)
  4. air
Answer and explanation

a vacuum (empty space)

Sound needs a medium of particles to travel through, so it cannot travel through a vacuum.

Problem 5

Using the words compression and rarefaction, explain how a speaker creates sound.
Answer and explanation

The speaker cone vibrates back and forth, pushing air molecules together to create compressions and pulling them apart to create rarefactions. This repeating pattern travels outward through the air as a sound wave.

Problem 6

A sound wave travels at 340 m/s and has a wavelength of 0.85 m. Find its frequency.
Answer and explanation

400 Hz

f = v ÷ λ = 340 ÷ 0.85 = 400 Hz.

Problem 7

An echo returns after 0.4 s. Using sound speed 340 m/s, find the distance to the reflecting wall. Explain the reasoning in your own words.
Answer and explanation

68 m to the wall in this model.

Total path=340×0.4=136 m.

The path goes to the wall and back, so wall distance=136/2=68 m.

Response criteria

  • Use the stated conditions.
  • Show the evidence or calculation.
  • Explain the conclusion and any stated limitation.

Problem 8

A sound has frequency 200 Hz and another 400 Hz. Which is higher pitched under the simple comparison? Show the reasoning, not just the conclusion.
Hints

Remember the return path; the echo delay is not one-way time.

Answer and explanation

The 400 Hz sound, because its frequency is higher.

Problem 9

Why cannot ordinary sound travel through a vacuum between two objects?
Hints

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

Answer and explanation

Sound needs particles of a medium to transmit mechanical disturbances; a vacuum lacks that medium.

Problem 10

Sound as compressions and rarefactionsSound is a pressure variation that needs a mediumcompressionrarefactiontravels through the mediumIn a vacuum there are no particles to compress, so sound does not travel
A sound wave is a pattern of crowding and spreading passing through matter. Drawing it as particles rather than as a curve makes the requirement obvious: remove the particles and there is nothing left to carry the pattern.
For this unit’s illustrated case, what travels from a speaker to the listener in this model?
Answer and explanation

The pressure disturbance and energy travel through the air. Individual particles move locally rather than carrying themselves from speaker to listener.

Draw a line of particles with alternating close and less-close spacing. Show travel and local motion along the same axis in the longitudinal model. Do not show one particle traveling all the way from a speaker to a listener.

The pressure disturbance and energy travel through the air. Individual particles move locally rather than carrying themselves from speaker to listener.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 11

An echo returns after 0.3 s in a supplied air model with sound speed 340 m/s. Find the reflector distance and explain why using the full 102 m as the one-way distance is wrong.
Answer and explanation

Round-trip length is 340×0.3 = 102 m, so reflector distance is 51 m. The measured time includes travel out and back.

Draw both travel legs.

Divide total path by two only under the stated same-path out-and-back comparison.

Round-trip length is 340×0.3 = 102 m, so reflector distance is 51 m. The measured time includes travel out and back.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 12

OpenSciEd graph A of source position in millimetres against time from 0 to 0.01 seconds, showing repeated vibrations around a mean position of about 20 millimetres.
OpenSciEd 8.2 Sound Waves, Lesson 5: actual position-versus-time record A. Read the labeled axes and estimate a cycle interval before opening the worked guidance. The vertical value is the vibrating point’s position relative to a coordinate origin, not air pressure or sound wavelength. Sound-source vibration graph A · OpenSciEd · 8.2 Sound Waves, Lesson 5, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.
Read OpenSciEd record A. Estimate a peak-to-peak cycle interval, calculate cycles per second and measure displacement amplitude about the midline. Explain why its mean position near 20 mm is not its amplitude. Approximate graph readings are expected.
Answer and explanation

One cycle takes about 0.0033 s, giving frequency about 300 Hz using cycles per second = 1/period. The midline is near 20 mm and extrema are about 19.2 and 20.8 mm, so displacement amplitude is about 0.8 mm. Amplitude measures departure from the midline, not distance from the coordinate origin. Allow reasonable approximate readings from this source graph.

Compare consecutive peaks: about 0.0033 s per cycle.

Compute 1/period, approximately 300 cycles per second, or Hz.

Amplitude is about 20.8−20 = 0.8 mm (half the peak-to-peak displacement).

Response criteria

  • Read time on the horizontal axis; do not call it wavelength.
  • Estimate frequency from repeats per second.
  • Measure amplitude from the midline and distinguish coordinate origin from equilibrium.

Problem 13

OpenSciEd graph B of source position in millimetres against time from 0 to 0.01 seconds, showing repeated vibrations around a mean position of about 20 millimetres.
OpenSciEd 8.2 Sound Waves, Lesson 5: actual position-versus-time record B. Read the labeled axes and estimate a cycle interval before opening the worked guidance. The vertical value is the vibrating point’s position relative to a coordinate origin, not air pressure or sound wavelength. Sound-source vibration graph B · OpenSciEd · 8.2 Sound Waves, Lesson 5, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.
Read OpenSciEd record B. Estimate its period, frequency and amplitude. Compare it with record A, whose frequency is about 300 Hz and amplitude about 0.8 mm. Which vibration property has changed by a factor of two, and which has stayed approximately the same?
Answer and explanation

One cycle takes about 0.0017 s, giving frequency about 600 Hz using cycles per second = 1/period. The midline is near 20 mm and extrema are about 19.2 and 20.8 mm, so displacement amplitude is about 0.8 mm. Amplitude measures departure from the midline, not distance from the coordinate origin. Allow reasonable approximate readings from this source graph. B has about twice A’s frequency and half its period, while displacement amplitude stays approximately 0.8 mm. Faster repetition does not automatically mean a larger vibration.

Compare consecutive peaks: about 0.0017 s per cycle.

Compute 1/period, approximately 600 cycles per second, or Hz.

Amplitude is about 20.8−20 = 0.8 mm (half the peak-to-peak displacement).

Response criteria

  • Read time on the horizontal axis; do not call it wavelength.
  • Estimate frequency from repeats per second.
  • Measure amplitude from the midline and distinguish coordinate origin from equilibrium.

Problem 14

OpenSciEd graph C of source position in millimetres against time from 0 to 0.01 seconds, showing repeated vibrations around a mean position of about 20 millimetres.
OpenSciEd 8.2 Sound Waves, Lesson 5: actual position-versus-time record C. Read the labeled axes and estimate a cycle interval before opening the worked guidance. The vertical value is the vibrating point’s position relative to a coordinate origin, not air pressure or sound wavelength. Sound-source vibration graph C · OpenSciEd · 8.2 Sound Waves, Lesson 5, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.
Read OpenSciEd record C. Estimate period, frequency and amplitude. The 0.010 s window contains about one and a half cycles. Why would counting only its one fully completed cycle underestimate the frequency? Use the peak-to-peak interval or include the partial cycle.
Answer and explanation

One cycle takes about 0.0067 s, giving frequency about 150 Hz using cycles per second = 1/period. The midline is near 20 mm and extrema are about 19.2 and 20.8 mm, so displacement amplitude is about 0.8 mm. Amplitude measures departure from the midline, not distance from the coordinate origin. Allow reasonable approximate readings from this source graph. About 1.5 cycles in 0.010 s gives 150 Hz. Counting only one ignores the half-cycle and would incorrectly give 100 Hz; use the measured period or account for the partial cycle.

Compare consecutive peaks: about 0.0067 s per cycle.

Compute 1/period, approximately 150 cycles per second, or Hz.

Amplitude is about 20.8−20 = 0.8 mm (half the peak-to-peak displacement).

Response criteria

  • Read time on the horizontal axis; do not call it wavelength.
  • Estimate frequency from repeats per second.
  • Measure amplitude from the midline and distinguish coordinate origin from equilibrium.

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