Sound • Topic 2 of 3

Characteristics of a Sound Wave

A sound wave can be described completely by a small set of measurable quantities. Although sound is longitudinal, it is convenient to draw it as a graph of pressure (or density) against distance — this looks like a transverse wave and lets us label its features clearly.

Wavelength ($\lambda$). The distance covered by one complete wave — for sound, the distance between two consecutive compressions (or two consecutive rarefactions). It is measured in metres ($\text{m}$) and represented by the Greek letter lambda, $\lambda$.

Frequency ($f$). The number of complete waves (oscillations) produced per second. Its SI unit is the hertz ($\text{Hz}$); $1\,\text{Hz}$ means one oscillation per second. Frequency is set by the source and does not change when the wave moves into a new medium.

Time period ($T$). The time taken to produce one complete wave. Frequency and time period are reciprocals: $T=\frac{1}{f}$. So a $50\,\text{Hz}$ source has a period of $0.02\,\text{s}$.

Amplitude. The maximum displacement of a particle from its rest position (or the maximum change in pressure/density). Amplitude decides the loudness — a bigger swing means a louder sound.

How we perceive these:

  • Pitch depends on frequency. High frequency means high pitch (a whistle); low frequency means low pitch (a drum). A woman's voice is usually higher-pitched than a man's.
  • Loudness depends on amplitude. A gently tapped drum is soft; a hard-struck drum is loud. Loudness is measured in decibels ($\text{dB}$).
  • Quality (timbre) lets us tell two sources apart even when they play the same note at the same loudness — that is why a flute and a violin sound different.

The wave equation. The speed of a sound wave links these quantities: $$v=f\lambda$$ where $v$ is speed ($\text{m/s}$), $f$ is frequency ($\text{Hz}$) and $\lambda$ is wavelength ($\text{m}$). Since speed is fixed in a given medium, a higher frequency must mean a shorter wavelength.

Speed in different media. Sound travels fastest in solids, slower in liquids and slowest in gases, because tightly packed particles pass on the disturbance more readily. In air at $25\,^\circ\text{C}$ the speed is about $346\,\text{m/s}$ (roughly $340\,\text{m/s}$ for calculations), in water about $1500\,\text{m/s}$ and in steel about $5960\,\text{m/s}$. The speed of sound also rises as temperature rises.

Range of hearing. A healthy human ear hears frequencies from about $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$ — the audible range. Sound below $20\,\text{Hz}$ is infrasound (produced by elephants, whales and earthquakes), and above $20{,}000\,\text{Hz}$ is ultrasound (used by bats, dolphins and in medical imaging).

Transverse representation of a sound wave showing wavelength and amplitudeWavelength, Amplitude and Frequencyrestwavelength (lambda)amplitudecrest = compressiontrough = rarefactionv = f x lambda T = 1 / fpitch depends on frequency, loudness on amplitude
1
Worked Example
A source produces a sound wave of frequency $500\,\text{Hz}$ that travels at $340\,\text{m/s}$ in air. Find its wavelength.
Solution
  1. Given: $f=500\,\text{Hz}$, $v=340\,\text{m/s}$. Required: wavelength $\lambda$.
  2. Use the wave equation $v=f\lambda$, so $\lambda=\dfrac{v}{f}$.
  3. Substitute: $\lambda=\dfrac{340}{500}$.
  4. Calculate: $\lambda=0.68\,\text{m}$.

Answer: The wavelength is $0.68\,\text{m}$.

2
Worked Example
The time period of a vibrating tuning fork is $0.004\,\text{s}$. Calculate its frequency.
Solution
  1. Given: time period $T=0.004\,\text{s}$. Required: frequency $f$.
  2. Frequency and period are reciprocals: $f=\dfrac{1}{T}$.
  3. Substitute: $f=\dfrac{1}{0.004}$.
  4. Calculate: $f=250\,\text{Hz}$.

Answer: The frequency of the tuning fork is $250\,\text{Hz}$.

3
Worked Example
A sound wave has a wavelength of $1.5\,\text{m}$ and a frequency of $220\,\text{Hz}$. Find the speed of sound in that medium.
Solution
  1. Given: $\lambda=1.5\,\text{m}$, $f=220\,\text{Hz}$. Required: speed $v$.
  2. Use $v=f\lambda$.
  3. Substitute: $v=220\times1.5$.
  4. Calculate: $v=330\,\text{m/s}$.

Answer: The speed of sound in the medium is $330\,\text{m/s}$.

4
Worked Example
In $0.5\,\text{s}$, a wave completes $40$ oscillations. Find its frequency and time period.
Solution
  1. Frequency = number of oscillations per second = $\dfrac{40}{0.5}$.
  2. So $f=80\,\text{Hz}$.
  3. Time period $T=\dfrac{1}{f}=\dfrac{1}{80}$.
  4. So $T=0.0125\,\text{s}$.

Answer: Frequency $=80\,\text{Hz}$ and time period $=0.0125\,\text{s}$.

5
Worked Example
Two sounds have the same loudness but a flute and a violin can still be told apart. Which characteristic of sound is responsible, and on what do pitch and loudness depend?
Solution
  1. Loudness depends on the amplitude of the wave; equal loudness means equal amplitude here.
  2. Pitch depends on the frequency of the wave.
  3. The feature that distinguishes two sources of the same pitch and loudness is the quality (timbre) of the sound.
  4. Timbre arises from the different mixture of frequencies each instrument produces.

Answer: Quality (timbre) lets us distinguish them; pitch depends on frequency and loudness depends on amplitude.

6
Worked Example
A bat emits an ultrasonic wave of frequency $50{,}000\,\text{Hz}$. If the speed of sound in air is $340\,\text{m/s}$, find its wavelength and state why humans cannot hear it.
Solution
  1. Given: $f=50{,}000\,\text{Hz}$, $v=340\,\text{m/s}$. Required: $\lambda$.
  2. Use $\lambda=\dfrac{v}{f}=\dfrac{340}{50000}$.
  3. Calculate: $\lambda=0.0068\,\text{m}=6.8\,\text{mm}$.
  4. The audible range for humans is $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$; $50{,}000\,\text{Hz}$ lies above it.

Answer: The wavelength is $0.0068\,\text{m}$ ($6.8\,\text{mm}$); humans cannot hear it because it is ultrasound, above the $20{,}000\,\text{Hz}$ upper limit.

Key Points

  • Wavelength $\lambda$ is the distance of one full wave (compression to compression); frequency $f$ is waves per second, in $\text{Hz}$.
  • Time period and frequency are reciprocals: $T=\frac{1}{f}$; amplitude is the maximum displacement from rest.
  • Pitch depends on frequency, loudness on amplitude, and quality (timbre) distinguishes sources of the same note.
  • The wave equation is $v=f\lambda$; sound travels fastest in solids, then liquids, then gases, and faster as temperature rises.
  • Audible range is $20\,\text{Hz}$–$20{,}000\,\text{Hz}$; below it is infrasound, above it is ultrasound.
Tap an option to check your answer0 / 4
Q1.The SI unit of frequency is the
Explanation: Frequency is measured in hertz ($\text{Hz}$); one hertz is one oscillation per second.
Q2.The pitch of a sound depends on its
Explanation: Higher frequency gives higher pitch; loudness, by contrast, depends on amplitude.
Q3.Using $v=f\lambda$, a $680\,\text{Hz}$ note travelling at $340\,\text{m/s}$ has wavelength
Explanation: $\lambda=v/f=340/680=0.5\,\text{m}$.
Q4.The human audible range of frequency is approximately
Explanation: A healthy human ear detects roughly $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$.