Week 13: Waves & Sound

\(\Delta\) ← if you see the word "Delta" instead of a triangle, please refresh the page! Thanks :)

Overview⚓︎

I recommend reviewing the following concepts, they will probably come up in recitation:

• Reading wave information from graphs

  • from a displacement vs time graph: amplitude, period, frequency
  • from a displacement vs position graph: amplitude, wavelength
  • being careful about what each axis means

• Wave speed on a string

  • \(v = \dfrac{\lambda}{T} = f\lambda\)
  • combining information from a time graph and a snapshot of the string
  • using fractions of a cycle or fractions of a wavelength when the graph does not show a full period directly

• Sound intensity and spherical spreading

  • \(I = \dfrac{P}{4\pi r^2}\) for a source radiating uniformly in all directions
  • intensity decreases like \(1/r^2\)
  • adding intensities from multiple independent sound sources

• Sound intensity level

  • \(\beta = 10\log_{10}\!\left(\dfrac{I}{I_0}\right)\)
  • \(I_0 = 1.0 \times 10^{-12}\ \mathrm{W/m^2}\)
  • equal sources add in intensity, not in decibels
  • for \(N\) identical sources: \(I_{\mathrm{tot}} = NI_{\mathrm{one}}\)

• Doppler effect

  • for a moving source and stationary observer:
  • approaching: \(f' = f\dfrac{v}{v-v_s}\)
  • receding: \(f' = f\dfrac{v}{v+v_s}\)
  • for air near room temperature: \(v \approx 331 + 0.60T_C \ \mathrm{m/s}\)

• Interference from two in-phase speakers

  • wavelength from \(\lambda = v/f\)
  • constructive interference: path difference \(= m\lambda\)
  • destructive interference: path difference \(= \left(m+\tfrac12\right)\lambda\)
  • on the line between the speakers, carefully writing each distance to the point

Practice Problems⚓︎

• If you're comfortable with the "★★★" problems, you should do great during recitation.
• To print these questions, simply press Ctrl + P while on this page, and it should come out formatted nicely -- just make sure to refresh first so all the math renders properly.
  • For a copy without solutions, click here.

Difficulty key:

• ★☆☆ = beginner
• ★★☆ = standard
• ★★★ = challenging / multi-step

[1] Wave Graphs and Speed on a String⚓︎

1. ★☆☆
   A displacement vs time graph is taken at one end of a string. The graph shows a maximum displacement of \(+3.0\ \mathrm{cm}\), a minimum displacement of \(-3.0\ \mathrm{cm}\), and the time between successive crests is \(0.40\ \mathrm{s}\).

   • (a) What is the amplitude?
   • (b) What is the period?
   • (c) What is the frequency?
   Solution

   The amplitude is the maximum displacement from equilibrium, so

   \[A = 3.0\ \mathrm{cm}.\]

   The time between successive crests is one period, so

   \[T = 0.40\ \mathrm{s}.\]

   The frequency is

   \[f = \frac{1}{T} = \frac{1}{0.40} = 2.5\ \mathrm{Hz}.\]

   
   Answer: (a) \(A=3.0\ \mathrm{cm}\); (b) \(T=0.40\ \mathrm{s}\); (c) \(f=2.5\ \mathrm{Hz}\).

2. ★☆☆
   A snapshot of a sinusoidal wave on a string shows a maximum displacement of \(2.5\ \mathrm{cm}\) and a distance of \(1.8\ \mathrm{m}\) between two neighboring crests.

   • (a) What is the amplitude?
   • (b) What is the wavelength?
   Solution

   The amplitude is the maximum displacement from equilibrium:

   \[A = 2.5\ \mathrm{cm}.\]

   The distance between neighboring crests is one wavelength, so

   \[\lambda = 1.8\ \mathrm{m}.\]

   
   Answer: (a) \(A=2.5\ \mathrm{cm}\); (b) \(\lambda=1.8\ \mathrm{m}\).

3. ★★☆
   At one point on a string, a displacement vs time graph shows that the wave has a period of \(0.25\ \mathrm{s}\). A snapshot of the string shows a wavelength of \(1.50\ \mathrm{m}\).

   • (a) Find the frequency.
   • (b) Find the wave speed.
   Solution

   The frequency is

   \[f = \frac{1}{T} = \frac{1}{0.25} = 4.0\ \mathrm{Hz}.\]

   Then the wave speed is

   \[v = f\lambda = (4.0)(1.50) = 6.0\ \mathrm{m/s}.\]

   
   Answer: (a) \(f=4.0\ \mathrm{Hz}\); (b) \(v=6.0\ \mathrm{m/s}\).

4. ★★☆
   A displacement vs time graph at a fixed point on a string shows troughs at \(t=0.10\ \mathrm{s}\) and \(t=0.46\ \mathrm{s}\). A snapshot of the string at one instant shows that four full wavelengths occupy \(2.40\ \mathrm{m}\) of the string.

   • (a) Find the period.
   • (b) Find the frequency.
   • (c) Find the wavelength.
   • (d) Find the wave speed.
   Solution

   The time between successive troughs is one period:

   \[T = 0.46 - 0.10 = 0.36\ \mathrm{s}.\]

   So the frequency is

   \[f = \frac{1}{T} = \frac{1}{0.36} = 2.78\ \mathrm{Hz}.\]

   If four full wavelengths occupy \(2.40\ \mathrm{m}\), then

   \[\lambda = \frac{2.40}{4} = 0.60\ \mathrm{m}.\]

   The wave speed is

   \[v = f\lambda = (2.78)(0.60) = 1.67\ \mathrm{m/s}.\]

   
   Answer: (a) \(T=0.36\ \mathrm{s}\); (b) \(f=2.78\ \mathrm{Hz}\); (c) \(\lambda=0.60\ \mathrm{m}\); (d) \(v=1.67\ \mathrm{m/s}\).

5. ★★★
   A point on a string is observed as the wave passes by. A crest occurs at \(t=0.18\ \mathrm{s}\), and the next time that same point passes through equilibrium moving downward is at \(t=0.27\ \mathrm{s}\). In a snapshot of the string taken at one instant, a crest is located at \(x=0.20\ \mathrm{m}\) and the nearest trough to its right is at \(x=0.80\ \mathrm{m}\).

   • (a) Find the period.
   • (b) Find the frequency.
   • (c) Find the wavelength.
   • (d) Find the wave speed.
   Solution

   From a crest to the next equilibrium crossing moving downward is one-quarter of a cycle, so

   \[\frac{T}{4} = 0.27 - 0.18 = 0.09\ \mathrm{s}.\]

   Therefore,

   \[T = 4(0.09) = 0.36\ \mathrm{s}.\]

   The frequency is

   \[f = \frac{1}{T} = \frac{1}{0.36} = 2.78\ \mathrm{Hz}.\]

   The distance from a crest to the nearest trough is half a wavelength:

   \[\frac{\lambda}{2} = 0.80 - 0.20 = 0.60\ \mathrm{m}.\]

   So

   \[\lambda = 1.20\ \mathrm{m}.\]

   The wave speed is

   \[v = f\lambda = (2.78)(1.20) = 3.33\ \mathrm{m/s}.\]

   
   Answer: (a) \(T=0.36\ \mathrm{s}\); (b) \(f=2.78\ \mathrm{Hz}\); (c) \(\lambda=1.20\ \mathrm{m}\); (d) \(v=3.33\ \mathrm{m/s}\).

6. ★★★
   A displacement vs time graph for a point on a string shows that the point goes from \(y=0\) moving upward to the next crest in \(0.060\ \mathrm{s}\). A snapshot of the string shows that the distance from one zero crossing with positive slope to the next zero crossing with positive slope is \(0.96\ \mathrm{m}\).

   • (a) Find the period.
   • (b) Find the frequency.
   • (c) Find the wavelength.
   • (d) Find the wave speed.
   • (e) How long does it take a crest to travel \(7.2\ \mathrm{m}\) along the string?
   Solution

   Going from a zero crossing moving upward to the next crest takes one-quarter of a cycle, so

   \[\frac{T}{4} = 0.060\ \mathrm{s}.\]

   Thus,

   \[T = 0.240\ \mathrm{s}.\]

   The frequency is

   \[f = \frac{1}{T} = \frac{1}{0.240} = 4.17\ \mathrm{Hz}.\]

   The distance from one zero crossing with positive slope to the next zero crossing with positive slope is one full wavelength, so

   \[\lambda = 0.96\ \mathrm{m}.\]

   The wave speed is

   \[v = f\lambda = (4.17)(0.96) = 4.00\ \mathrm{m/s}.\]

   The travel time for a crest to move \(7.2\ \mathrm{m}\) is

   \[t = \frac{d}{v} = \frac{7.2}{4.00} = 1.8\ \mathrm{s}.\]

   
   Answer: (a) \(T=0.240\ \mathrm{s}\); (b) \(f=4.17\ \mathrm{Hz}\); (c) \(\lambda=0.96\ \mathrm{m}\); (d) \(v=4.00\ \mathrm{m/s}\); (e) \(t=1.8\ \mathrm{s}\).

[2] Sound Intensity and Sound Level⚓︎

1. ★☆☆
   A small speaker emits sound uniformly in all directions with a power output of \(18.0\ \mathrm{W}\).

   • What is the sound intensity at a distance of \(6.00\ \mathrm{m}\) from the speaker?
   Solution

   For a source radiating uniformly in all directions,

   \[I = \frac{P}{4\pi r^2}.\]

   Substituting,

   \[I = \frac{18.0}{4\pi(6.00)^2} = 3.98\times 10^{-2}\ \mathrm{W/m^2}.\]

   
   Answer: \(I = 3.98\times 10^{-2}\ \mathrm{W/m^2}\).

2. ★☆☆
   A sound wave has an intensity level of \(72.0\ \mathrm{dB}\).

   • What is its intensity?
   Solution

   The intensity level is related to intensity by

   \[\beta = 10\log_{10}\!\left(\frac{I}{I_0}\right).\]

   Solving for \(I\),

   \[I = I_0\,10^{\beta/10}.\]

   With \(I_0 = 1.0\times10^{-12}\ \mathrm{W/m^2}\),

   \[I = \left(1.0\times10^{-12}\right)10^{7.2} = 1.58\times10^{-5}\ \mathrm{W/m^2}.\]

   
   Answer: \(I = 1.58\times10^{-5}\ \mathrm{W/m^2}\).

3. ★★☆
   At your location, one room fan produces a sound intensity level of \(79.0\ \mathrm{dB}\). How many identical fans would be needed to produce a sound intensity level of \(88.0\ \mathrm{dB}\) at that same location?

   Solution

   For identical independent sources, intensities add:

   \[I_{\text{tot}} = NI_1.\]

   In decibels,

   \[\beta_{\text{tot}} - \beta_1 = 10\log_{10}(N).\]

   Therefore,

   \[88.0 - 79.0 = 10\log_{10}(N).\]

   So

   \[9.0 = 10\log_{10}(N)\]

   \[\log_{10}(N)=0.90\]

   \[N = 10^{0.90} = 7.94.\]

   Since you need a whole number of fans, you need

   \[N = 8.\]

   
   Answer: \(8\) fans.

4. ★★☆
   One student speaking alone produces a sound intensity level of \(56.0\ \mathrm{dB}\) at the back of a classroom. If \(12\) students are all talking independently with the same intensity at that same location, what sound intensity level results?

   Solution

   For \(12\) identical sound sources,

   \[I_{\text{tot}} = 12I_1.\]

   So the new sound level is

   \[\beta_{\text{tot}} = 10\log_{10}\!\left(\frac{12I_1}{I_0}\right).\]

   This can be written as

   \[\beta_{\text{tot}} = 10\log_{10}(12) + 10\log_{10}\!\left(\frac{I_1}{I_0}\right).\]

   Therefore,

   \[\beta_{\text{tot}} = 10\log_{10}(12) + 56.0.\]

   Since \(10\log_{10}(12)=10.79\),

   \[\beta_{\text{tot}} = 66.8\ \mathrm{dB}.\]

   
   Answer: \(66.8\ \mathrm{dB}\).

5. ★★★
   Two identical outdoor speakers are separated by \(24.0\ \mathrm{m}\). Each emits sound uniformly in all directions with a power output of \(0.400\ \mathrm{W}\).

   • (a) Find the total sound intensity at the midpoint between the speakers.
   • (b) A person then walks \(5.00\ \mathrm{m}\) directly toward one of the speakers. What total sound intensity do they hear there?
   Solution

   At the midpoint, the person is

   \[r = \frac{24.0}{2} = 12.0\ \mathrm{m}\]

   from each speaker.

   The intensity from one speaker is

   \[I_1 = \frac{P}{4\pi r^2} = \frac{0.400}{4\pi(12.0)^2}.\]

   Since there are two identical speakers,

   \[I_{\text{mid}} = 2I_1 = \frac{0.800}{4\pi(12.0)^2} = 4.42\times10^{-4}\ \mathrm{W/m^2}.\]

   After walking \(5.00\ \mathrm{m}\) toward one speaker, the distances are

   \[r_{\text{near}} = 12.0 - 5.0 = 7.0\ \mathrm{m}\]

   and

   \[r_{\text{far}} = 12.0 + 5.0 = 17.0\ \mathrm{m}.\]

   So the total intensity is

   \[I_{\text{tot}} = \frac{0.400}{4\pi(7.0)^2} + \frac{0.400}{4\pi(17.0)^2}.\]

   Numerically,

   \[I_{\text{tot}} = 7.60\times10^{-4}\ \mathrm{W/m^2}.\]

   
   Answer: (a) \(4.42\times10^{-4}\ \mathrm{W/m^2}\); (b) \(7.60\times10^{-4}\ \mathrm{W/m^2}\).

6. ★★★
   Thirty-six identical machines together produce a sound intensity level of \(96.0\ \mathrm{dB}\) at a certain point in a workshop.

   • (a) Find the total sound intensity produced by all \(36\) machines.
   • (b) Find the intensity produced by one machine.
   • (c) What sound intensity level would result if only \(4\) of these machines were running?
   Solution

   The total intensity corresponding to \(96.0\ \mathrm{dB}\) is

   \[I_{\text{tot}} = I_0\,10^{\beta/10} = \left(1.0\times10^{-12}\right)10^{9.6} = 3.98\times10^{-3}\ \mathrm{W/m^2}.\]

   Since there are \(36\) identical machines,

   \[I_1 = \frac{I_{\text{tot}}}{36} = \frac{3.98\times10^{-3}}{36} = 1.11\times10^{-4}\ \mathrm{W/m^2}.\]

   If only \(4\) machines are running, the intensity is

   \[I_4 = 4I_1 = 4.42\times10^{-4}\ \mathrm{W/m^2}.\]

   The corresponding sound intensity level is

   \[\beta_4 = 10\log_{10}\!\left(\frac{I_4}{I_0}\right)\]

   \[\beta_4 = 10\log_{10}\!\left(\frac{4.42\times10^{-4}}{1.0\times10^{-12}}\right) = 86.5\ \mathrm{dB}.\]

   
   Answer: (a) \(3.98\times10^{-3}\ \mathrm{W/m^2}\); (b) \(1.11\times10^{-4}\ \mathrm{W/m^2}\); (c) \(86.5\ \mathrm{dB}\).

[3] Doppler Effect and Two-Speaker Interference⚓︎

1. ★☆☆
   What is the speed of sound in air on a day when the temperature is \(22.0^\circ\mathrm{C}\)?

   Solution

   Use

   \[v \approx 331 + 0.60T_C.\]

   With \(T_C = 22.0\),

   \[v = 331 + 0.60(22.0) = 344.2\ \mathrm{m/s}.\]

   
   Answer: \(v = 344.2\ \mathrm{m/s}\).

2. ★☆☆
   A sound wave in air has frequency \(680\ \mathrm{Hz}\). If the speed of sound is \(340\ \mathrm{m/s}\), what is the wavelength?

   Solution

   The wavelength is

   \[\lambda = \frac{v}{f} = \frac{340}{680} = 0.500\ \mathrm{m}.\]

   
   Answer: \(\lambda = 0.500\ \mathrm{m}\).

3. ★★☆
   A police car moves at \(31.0\ \mathrm{m/s}\) while sounding a siren of frequency \(720\ \mathrm{Hz}\). The air temperature is \(25.0^\circ\mathrm{C}\).

   • What frequency is heard by a stationary observer when the police car is approaching?
   Solution

   First find the speed of sound:

   \[v = 331 + 0.60(25.0) = 346\ \mathrm{m/s}.\]

   For a moving source approaching a stationary observer,

   \[f' = f\frac{v}{v-v_s}.\]

   Therefore,

   \[f' = 720\frac{346}{346-31.0} = 720\frac{346}{315} = 791\ \mathrm{Hz}.\]

   
   Answer: \(f' \approx 791\ \mathrm{Hz}\).

4. ★★☆
   Two in-phase loudspeakers, \(A\) and \(B\), face each other and are separated by \(1.50\ \mathrm{m}\). They emit sound of frequency \(680\ \mathrm{Hz}\), and the speed of sound is \(340\ \mathrm{m/s}\).

   • (a) What is the wavelength?
   • (b) Locate all points on the line between the speakers, measured from speaker \(A\), where destructive interference occurs.
   Solution

   The wavelength is

   \[\lambda = \frac{v}{f} = \frac{340}{680} = 0.500\ \mathrm{m}.\]

   Let \(x\) be the distance from speaker \(A\). Then the distances to the two speakers are

   \[r_A = x, \qquad r_B = 1.50 - x.\]

   The path difference is

   \[\Delta r = |r_B-r_A| = |1.50 - 2x|.\]

   For destructive interference,

   \[\Delta r = \left(m+\frac12\right)\lambda.\]

   Since \(\lambda = 0.500\ \mathrm{m}\), the allowed path differences between the speakers are

   \[0.25,\ 0.75,\ 1.25\ \mathrm{m}.\]

   Solve each case:

   \[|1.50 - 2x| = 0.25 \quad \Rightarrow \quad x=0.625,\ 0.875\ \mathrm{m}\]

   \[|1.50 - 2x| = 0.75 \quad \Rightarrow \quad x=0.375,\ 1.125\ \mathrm{m}\]

   \[|1.50 - 2x| = 1.25 \quad \Rightarrow \quad x=0.125,\ 1.375\ \mathrm{m}\]

   
   Answer: (a) \(\lambda = 0.500\ \mathrm{m}\); (b) destructive interference at \(x = 0.125,\ 0.375,\ 0.625,\ 0.875,\ 1.125,\ 1.375\ \mathrm{m}\) from speaker \(A\).

5. ★★★
   An ambulance travels at \(27.0\ \mathrm{m/s}\) and emits a siren of frequency \(950\ \mathrm{Hz}\). The air temperature is \(15.0^\circ\mathrm{C}\).

   • (a) Find the speed of sound in air.
   • (b) What frequency is heard by a stationary observer as the ambulance approaches?
   • (c) What frequency is heard by the observer after the ambulance passes and is moving away?
   Solution

   First find the speed of sound:

   \[v = 331 + 0.60(15.0) = 340\ \mathrm{m/s}.\]

   For the approaching case,

   \[f'_{\text{app}} = f\frac{v}{v-v_s} = 950\frac{340}{340-27.0} = 950\frac{340}{313} = 1.03\times10^3\ \mathrm{Hz}.\]

   So

   \[f'_{\text{app}} \approx 1030\ \mathrm{Hz}.\]

   For the receding case,

   \[f'_{\text{away}} = f\frac{v}{v+v_s} = 950\frac{340}{340+27.0} = 950\frac{340}{367} = 879\ \mathrm{Hz}.\]

   
   Answer: (a) \(v=340\ \mathrm{m/s}\); (b) \(f' \approx 1030\ \mathrm{Hz}\) approaching; (c) \(f' \approx 879\ \mathrm{Hz}\) receding.

6. ★★★
   Two in-phase loudspeakers, \(A\) and \(B\), face each other and are separated by \(1.50\ \mathrm{m}\). They emit sound of frequency \(660\ \mathrm{Hz}\), and the speed of sound is \(330\ \mathrm{m/s}\).

   • (a) What is the wavelength?
   • (b) Locate all points on the line between the speakers, measured from speaker \(A\), where constructive interference occurs.
   Solution

   The wavelength is

   \[\lambda = \frac{v}{f} = \frac{330}{660} = 0.500\ \mathrm{m}.\]

   Let \(x\) be the distance from speaker \(A\). Then

   \[r_A = x, \qquad r_B = 1.50 - x.\]

   The path difference is

   \[\Delta r = |r_B-r_A| = |1.50 - 2x|.\]

   For constructive interference,

   \[\Delta r = m\lambda.\]

   Since \(\lambda = 0.500\ \mathrm{m}\), the allowed path differences between the speakers are

   \[0,\ 0.50,\ 1.00,\ 1.50\ \mathrm{m}.\]

   The \(1.50\ \mathrm{m}\) case corresponds to the speaker locations themselves, so for interior points we use

   \[0,\ 0.50,\ 1.00\ \mathrm{m}.\]

   Solve each case:

   \[|1.50 - 2x| = 0 \quad \Rightarrow \quad x=0.750\ \mathrm{m}\]

   \[|1.50 - 2x| = 0.50 \quad \Rightarrow \quad x=0.500,\ 1.000\ \mathrm{m}\]

   \[|1.50 - 2x| = 1.00 \quad \Rightarrow \quad x=0.250,\ 1.250\ \mathrm{m}\]

   
   Answer: (a) \(\lambda = 0.500\ \mathrm{m}\); (b) constructive interference at \(x = 0.250,\ 0.500,\ 0.750,\ 1.000,\ 1.250\ \mathrm{m}\) from speaker \(A\).