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E3.5 · Relate sound speed to the particle nature of a medium

Learn to relate sound speed to the particle nature of a medium through clear examples and targeted practice.

Ontario Grade 11 Physics

Waves and Sound

Ontario Grade 11 Physics — E3.5

Sound is a vibration that travels through matter. The matter carrying the sound is called the medium. Air, water, and steel are examples of media. Sound does not travel in empty space because there are no particles there to pass on the vibration. In this lesson, we connect sound speed to the way particles are arranged and interact. We also use the course-level wave relationship between speed, frequency, and wavelength.

What you will learn

1. Prerequisite bridge: particles and waves

A particle is a small piece of matter, such as an atom or molecule. A solid, liquid, or gas is made of particles. The particles in a solid stay near fixed positions. They can vibrate. In a liquid, particles remain close but can move around one another. In a gas, particles are much farther apart and move freely.
A vibration is repeated back-and-forth motion. A sound wave begins when a source vibrates. Nearby particles vibrate too, and their interactions pass the disturbance through the medium. The particles do not travel all the way from the source to the listener. Instead, the disturbance and its energy move through the medium.
Frequency is the number of complete vibrations each second. Its SI unit is the hertz, written HzHz (one hertz means one vibration per second). Wavelength is the distance between matching points on neighbouring waves, such as one compression to the next. Its SI unit is the metre, m\mathrm{m}. Sound speed is the distance the disturbance travels each second. It is measured in metres per second, m/s\mathrm{m/s}.
v=fλv=f\lambda

2. How particle nature affects sound speed

The physical system is the sound disturbance and the particles in the medium through which it travels. Sound speed is a scalar: it has a size but no direction. The direction of travel is described separately. For a one-dimensional description, choose the direction from the source toward the listener as positive. This choice does not change the speed.
Particles pass on a sound disturbance through their interactions with nearby particles. When particles are close and interact effectively, the disturbance can be passed along quickly. In general, sound travels faster in solids than in gases. Many solids have particles close together and strong interactions. Gas particles are farther apart, so the disturbance is passed from particle to particle less quickly. Liquids commonly have sound speeds between those of gases and solids, but the exact speed depends on the particular medium.
This is a general pattern, not a rule that lets us calculate every speed from particle spacing alone. Different substances have different particle arrangements and interactions. Their sound speeds must be known or supplied when a numerical calculation is needed. Temperature can also affect sound speed, especially in a gas, so comparisons should specify the conditions when relevant.
A wave diagram can show the direction of travel without suggesting that particles move along with the wave. The dots represent particles. Each particle vibrates near its usual position as the disturbance advances.

3. Using the wave model to compare media

The relationship v=fλv=f\lambda connects sound speed vv, frequency ff, and wavelength λ\lambda. Frequency is set by the vibrating source. When the sound enters a different medium, its speed can change because the particles and their interactions differ. For a sound of the same frequency, the wavelength must change with the speed.
Before calculating, identify the medium and the known quantity. Keep units with each value. The frequency in hertz is equivalent to inverse seconds, so multiplying frequency by wavelength gives metres per second. Report a suitable number of significant figures. If a direction is requested, state the propagation direction using the chosen positive direction; do not attach a direction to the scalar speed.
A result should fit the situation. A wavelength of a few metres can be reasonable for a low-frequency sound, while a much shorter wavelength can result from a higher frequency or lower sound speed. A calculation does not prove a value was measured; it uses a supplied value and a model.
λ=vf\lambda=\frac{v}{f}

Worked example

Finding wavelength in air

A sound in air has a frequency of 500 Hz500\,\mathrm{Hz}. Use a sound speed of 343 m/s343\,\mathrm{m/s}. Find its wavelength. Take the positive direction to be from the source toward the listener.
  1. Define the quantities
    The system is the sound travelling through air. Its propagation direction is positive, from source to listener. The speed is a scalar, while the direction describes where the sound travels. The unknown is wavelength.
    f=500 Hz,v=343 m/sf=500\,\mathrm{Hz},\quad v=343\,\mathrm{m/s}
  2. Choose the relationship
    The wave relationship connects speed, frequency, and wavelength. Rearranging it isolates the unknown wavelength.
    λ=vf\lambda=\frac{v}{f}
  3. Substitute and calculate
    Substitute the supplied values with their units. Round the result to three significant figures, matching the given values.
    λ=343 m/s500 s−1=0.686 m\lambda=\frac{343\,\mathrm{m/s}}{500\,\mathrm{s^{-1}}}=0.686\,\mathrm{m}
Answer: The wavelength is 0.686 m0.686\,\mathrm{m}. The sound travels in the positive direction, from source to listener.
Check: The units reduce to metres because (m/s)/(s−1)=m(\mathrm{m/s})/(\mathrm{s^{-1}})=\mathrm{m}. A wavelength under one metre is reasonable for a 500 Hz sound in air. The wavelength is positive because it is a distance.

Worked example

Comparing the same sound in air and water

A source produces a 740 Hz740\,\mathrm{Hz} sound. Use 343 m/s343\,\mathrm{m/s} for air and 1480 m/s1480\,\mathrm{m/s} for water. Find the wavelength in each medium and compare them. The sound travels from the source toward the listener.
  1. Identify the system and known values
    The system is the sound in each medium. The positive direction is from source to listener. Frequency stays the same for this comparison, while the supplied medium-dependent speeds differ.
    f=740 Hz,vair=343 m/s,vwater=1480 m/sf=740\,\mathrm{Hz},\quad v_{\mathrm{air}}=343\,\mathrm{m/s},\quad v_{\mathrm{water}}=1480\,\mathrm{m/s}
  2. Calculate each wavelength
    Use the same relationship for both media. Keeping the frequency fixed makes the wavelength comparison show how the supplied speeds affect the wave.
    λair=343740 m=0.464 m,λwater=1480740 m=2.00 m\lambda_{\mathrm{air}}=\frac{343}{740}\,\mathrm{m}=0.464\,\mathrm{m},\quad \lambda_{\mathrm{water}}=\frac{1480}{740}\,\mathrm{m}=2.00\,\mathrm{m}
  3. Compare and interpret
    The water wavelength is longer because the supplied sound speed in water is greater. The faster speed is consistent with the general pattern that sound travels faster in many liquids than in gases, due to differences in particle arrangement and interactions.
    2.00 m0.464 m≈4.31\frac{2.00\,\mathrm{m}}{0.464\,\mathrm{m}}\approx4.31
Answer: The wavelengths are 0.464 m0.464\,\mathrm{m} in air and 2.00 m2.00\,\mathrm{m} in water. The water wavelength is about 4.314.31 times as long.
Check: Each calculation gives metres, and both wavelengths are positive. Multiplying each wavelength by 740 Hz740\,\mathrm{Hz} returns the supplied speed in m/s\mathrm{m/s}. The longer water wavelength agrees with its greater supplied speed at the same frequency.

Worked example

Finding speed from a wavelength

In a solid, a sound has a wavelength of 3.2 m3.2\,\mathrm{m} and a frequency of 1200 Hz1200\,\mathrm{Hz}. Find its speed. The sound travels in the positive direction from the source toward the listener.
  1. Set the system and unknown
    The system is the sound disturbance in the solid. Its direction is positive, from source to listener. The unknown is the scalar speed; the propagation direction is not part of the speed value.
    λ=3.2 m,f=1200 Hz\lambda=3.2\,\mathrm{m},\quad f=1200\,\mathrm{Hz}
  2. Apply the wave relationship
    Speed is frequency multiplied by wavelength. This is suitable because both quantities are supplied.
    v=fλv=f\lambda
  3. Substitute with units
    Multiply the frequency by the wavelength. The given wavelength has two significant figures, so report the speed to two significant figures.
    v=(1200 s−1)(3.2 m)=3.8×103 m/sv=(1200\,\mathrm{s^{-1}})(3.2\,\mathrm{m})=3.8\times10^3\,\mathrm{m/s}
Answer: The sound speed is 3.8×103 m/s3.8\times10^3\,\mathrm{m/s} in the positive direction.
Check: The units are s−1×m=m/s\mathrm{s^{-1}}\times\mathrm{m}=\mathrm{m/s}. This speed is much greater than typical sound speed in air, which is reasonable for a sound travelling through a solid.

Common mistakes and how to avoid them

Thinking that particles travel from the source to the listener along with the sound.
Correction: Particles vibrate near their usual positions. The disturbance passes through the medium.
Treating sound speed as the same in every material.
Correction: Sound speed depends on the medium's particles and their interactions. Use the value for the specified medium.
Assuming a higher frequency always means a higher sound speed.
Correction: Frequency alone does not set the speed. The medium matters. For a given speed, a higher frequency corresponds to a shorter wavelength.
Giving a direction as part of a speed value.
Correction: Speed is a scalar. State the direction of sound propagation separately when it is needed.

Lesson summary

Check your understanding

Question 1

Why can sound travel through a solid but not through empty space?
  1. A solid has particles that can pass on the vibration; empty space has no particles.
  2. Sound requires visible light to guide it.
  3. Sound travels only when particles move all the way from source to listener.
  4. correctIndex} 0
Show answer and explanation
A solid has particles that can pass on the vibration; empty space has no particles.
Sound is passed through interactions between particles. Empty space has no particles to pass on the disturbance.

Question 2

A sound has frequency 400 Hz400\,\mathrm{Hz} and speed 320 m/s320\,\mathrm{m/s}. What is its wavelength?
  1. 0.80 m0.80\,\mathrm{m}
  2. 1.25 m1.25\,\mathrm{m}
  3. 128000 m128000\,\mathrm{m}
  4. correctIndex} 0
Show answer and explanation
0.80 m0.80\,\mathrm{m}
Using λ=v/f\lambda=v/f gives 320/400=0.80 m320/400=0.80\,\mathrm{m}. The units reduce to metres.

Question 3

The same sound frequency travels through two media. The sound speed is greater in medium A. Which statement is correct?
  1. Its wavelength is greater in medium A.
  2. Its wavelength is smaller in medium A.
  3. Its wavelength must be unchanged because frequency is unchanged.
  4. correctIndex} 0
Show answer and explanation
Its wavelength is greater in medium A.
From λ=v/f\lambda=v/f, if frequency is unchanged, greater speed means greater wavelength.

Key terms

Medium
Matter through which a wave travels.
Particle
A small piece of matter, such as an atom or molecule.
Vibration
Repeated back-and-forth motion.
Frequency
The number of complete vibrations each second, measured in hertz.
Wavelength
The distance between matching points on neighbouring waves.
Sound speed
The distance the sound disturbance travels each second, measured in metres per second.
Scalar
A quantity with size but no direction.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation E3.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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