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chapter_14

Ch.14: Waves and sound (Apr 3-10)

Concepts Ch.14

Units Ch.14

Video 2.1: Ruben's tube, standing wave with explanation. Video 2.2: Standing waves on a 2D surface.

Constants Ch.14

From: Walker, James S. “Ch. 14 Waves and Sound.” Physics. Upper Saddle River, NJ: Pearson/Prentice Hall, 2004

Table 14.1: Speed of sound in various materials
Material Temperature Speed of sound (m/s)
Aluminum 6420
Granite 6000
Steel 5960
Pyrex glass 5640
Copper 5010
Plastic 2680

Fresh water

20 $^\circ$C 1482
0 $^\circ$C 1402
Hydrogen 0 $^\circ$C 1284
Helium 0 $^\circ$C 965

Air

20 $^\circ$C 343
0 $^\circ$C 331

Math Ch.14

Addition of trig functions Doppler equation approximation for slow observer/source speeds $u\ll v$
$\sin\alpha+\sin\beta=$ $2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$ $\left(1+\frac{u}{v}\right)^2$ $\approx 1+2\frac{u}{v}$ $\left(1-\frac{u}{v}\right)^2$ $\approx 1-2\frac{u}{v}$
$\cos\alpha+\cos\beta=$ $2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$ $\left(1+\frac{u}{v}\right)\left(1-\frac{u}{v}\right)$ $\,=\, 1-\left(\frac{u}{v}\right)^2$ $\,\approx\, 1$ $\left(1+\frac{u_1}{v}\right)\left(1-\frac{u_2}{v}\right)$ $\approx 1+\frac{u_1-u_2}{v}$
$\cos\alpha+\sin\beta=$ $-2\sin\left(\frac{\alpha+\beta}{2}+\frac{\pi}{4}\right)\sin\left(\frac{\alpha-\beta}{2}-\frac{\pi}{4}\right)$ $\left(\frac{1}{1+\frac{u}{v}}\right)$ $\approx 1-\frac{u}{v}$ $\left(\frac{1}{1-\frac{u}{v}}\right)$ $\approx 1+\frac{u}{v}$
$\cos\alpha-\cos\beta=$ $-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$ $\left(\frac{1+\frac{u}{v}}{1-\frac{u}{v}}\right)$ $\approx 1+2\frac{u}{v}$ $\left(\frac{1-\frac{u}{v}}{1+\frac{u}{v}}\right)$ $\approx 1-2\frac{u}{v}$ $\left(\frac{1+\frac{u_1}{v}}{1-\frac{u_2}{v}}\right)$ $\approx 1+\frac{u_1+u_2}{v}$

Lecture notes Apr 3-10

Equation Sheet Ch.14

Please update if an equation is not included

  1. Definition of wavelength ($\lambda$):
    • The distance over which a wave repeats itself.
      • SI unit: meter (m)
  2. Period ($T$) of a wave:
    • Time required for one wavelength to pass a given point.
      • SI unit: second (s)
  3. Frequency $(f)$ of a wave:
    • $f =\frac{1}{T}$
      • SI unit: hertz (Hz)
  4. Speed of a wave ($v$)
    • $v=\frac{\text{distance}}{\text{time}}$ $= \frac{\lambda}{T} = \lambda\cdot f$
      • SI unit: meter per second (m/s)
  5. Speed of a wave on a string, $v$, with a mass per length $\mu$, and tension, $F_T$:
    • $v = \sqrt\frac{F_T}{\mu}$
      • SI unit: m/s.
    • $\mu=\frac{m}{L}$ is the mass of the string per unit length (linear mass density)
      • SI unit: kg/m.
    • $F_T$ is the tension force
      • SI unit: newton (N = $\frac{\text{kg}\cdot{\text m}}{ {\text s}^2}$)
  6. Wave function of a harmonic wave of wavelength $\lambda$ and period $T$ is described by:
    • $y(x,t) =$ $A\cos\big(\frac{2\pi}{\lambda}x \mp \frac{2\pi}{T}t+\theta_0\big)$
      • if the x and t terms inside the cos have the opposite sign, the wave is moving in the $+x$ direction
      • if the x and t terms inside the cos have the same sign, the wave is moving in the $-x$ direction
      • $\frac{2\pi}{T}=\omega$ is the angular frequency
        • SI unit: rad/s or s-1
      • $\frac{2\pi}{\lambda}=k$ is the wavenumber (not to be confused with the spring constant)
        • SI unit: m-1
      • In a generic wave equation, e.g. $y=$ $- 0.2\sin\big(2x-3t+\frac{\pi}{4}\!\big)\;$ or $\;y=$ $- 0.2\cos\big(2x-3t-\frac{\pi}{4}\!\big)$
        • The amplitude here is 0.2 (always positive), no matter sin or cos.
          • The sign in front of the amplitude (in this case, $-$) just shifts the wave by half-period (or half-wavelength)
          • Changing sin to cos or vice versa, or changing the phase term (here, $\frac{\pi}{4}$ or $-\frac{\pi}{4}$) just shifts the wave in time and/or space, but does not affect its amplitude, frequency, wavelength, speed, etc…
        • The wavelength (always positive) can be found from the magnitude of coefficient in front of $x$, in this case: $\big|2\big|=\frac{2\pi}{\lambda}$
          • $\lambda=\frac{2\pi}{|2|}$ $=\frac{1}{\pi}$ $\approx 0.32\,$m
        • The period (always positive) can be found from the magnitude of the coefficient in front of $t$, in this case: $\big|\!-\!3\big|=\frac{2\pi}{T}$
          • $T=\frac{2\pi}{|-3|}$ $=\frac{2\pi}{3}$ $\approx 2.1\,$s
        • The frequency (always positive) can be found from the magnitude of the coefficient in front of $t$, in this case: $\big|\!-\!3\big|=2\pi f$
          • $f=\frac{|-3|}{2\pi}$ $=\frac{3}{2\pi}$ $\approx 0.48\,$Hz
        • The speed of the wave can be found from the negative of the ratio of the t-coefficient to the x-coefficient:
          • $v=-\frac{-3}{2}$ $=+1.5\frac{\text m}{\text s}$ (in this case, in $+x$ direction, also assuming the equation is written using SI units)
  7. Definition of intensity: transmitted power per unit area of the receiver, where power is energy per unit time
    • $I = \frac{P}{A}$
      • SI unit: $\frac{\text W}{ {\text m}^2}$
  8. Intensity of sound from a point source of acoustic power $P$, at a distance $r$ from the source:
    • $I = \frac{P}{4πr^2}$
      • Here the surface area of a sphere ($A=4\pi r^2$) is used because the sound waves from a point source radiate outward in three-dimensional space, each wavefront forming a sphere of an ever-increasing radius r.
      • SI unit: $\frac{\text W}{ {\text m}^2}$
  9. Definition of intensity level:
    • $\beta = 10\,{\text{dB}}\cdot\log\left(\frac{I}{I_0}\right)$
      • SI unit: decibel (dB, dimensionless).
    • $I_0=10^{-12}\frac{\text W}{ {\text m}^2}$ is the absolute threshold of hearing
    • log is the decimal log, such that
      • log(10)=1
      • log(100)=2
      • log(1000)=3, etc. Not to be confused with the natural logarithm
  10. Doppler Effect for moving observer
    • Apparent frequency of sound: $\;\;f' = \left(1 \pm \frac{u_o}{v}\right)\,f$
      • SI unit: Hz $=\frac{1}{\text s} = {\text s}^{-1}$
    • $u_o$ is the speed of observer
    • $v$ is the wave speed
    • Plus sign is used when the observer moves toward the source and minus sign when the observer moves away from the source.
  11. Doppler Effect for moving sound source:
    • Apparent frequency of sound: $\;\;f' = \big(\!\frac{1}{1 \mp \frac{u_s}{v}}\!\big)\,f$
      • SI unit: Hz $=\frac{1}{\text s} = {\text s}^{-1}$
    • $u_s$ is the speed of the source
    • $v$ is the wave speed
    • Minus sign is used when the source moves toward the observer and the plus sign when the source moves away from the observer.
  12. Doppler Effect for moving sound source and moving observer:
    • $f' = \frac{1 \pm \frac{u_o}{v}}{1 \mp \frac{u_s}{v}}f$
      • SI unit: Hz $=\frac{1}{\text s} = {\text s}^{-1}$
    • In the numerator, the plus sign corresponds to the case in which the observer moves in the direction of the source, whereas the minus sign indicates motion in the opposite direction.
    • In the denominator, the minus sign corresponds to the case in which the source moves in the direction of the observer, whereas the plus sign indicates motion in the opposite direction.
  13. For standing waves on a string, or in the column of air, when both ends are of the same type:
    • Wavelength of the nth harmonic:
      • $\lambda_n = \frac{\lambda_1}{n}$ $= \frac{2L}{n}$
    • frequency of the nth harmonic:
      • $f_n = n\,f_1$ $= n\!\frac{v}{2L}$
    • in both expressions above
      • $n=1,2,3,4,5...$ is a positive integer
    • v is the speed of the wave:
      • $v=\sqrt{\frac{F_T}{\mu}}$ for the string
      • speed of sound for the column of air
  14. For standing waves on a string, or in the column of air, when both ends are of the opposite type:
    • Wavelength of the nth harmonic:
      • $\lambda_n = \frac{\lambda_1}{n}$ $= \frac{4L}{n}$
    • frequency of the nth harmonic:
      • $f_n = n\,f_1$ $= n\!\frac{v}{4L}$
    • in both expressions above
      • $n=1,3,5...$ is an odd positive integer
  15. Definition of Beat Frequency (the difference between the individual frequencies):
    • $f_\text{beat} = \big|\,f_1-f_2\big|$
      • SI unit: Hz $=\frac{1}{\text s} = {\text s}^{-1}$

Examples Ch.14


Problem 2.1

What is the wavelength of the lowest and the highest frequency you can hear?

Steps:

  1. Determine (or look up) the range of audible frequencies
    • my lower range is $f_1=35\,$Hz (some people can hear down to $f_\text{min}=20\,$Hz)
    • my higher range is $f_2=15.5\,$kHz (some people can hear up to $f_\text{max}=20\,$kHz)
  2. Look up the speed of sound at room temperature
    • $v=343\,\frac{\text m}{\text s}$
  3. Calculate the wavelength from $v=\lambda\cdot f$:
    `
    • $\lambda_{f{\text{max}}}=\frac{v}{f_\text{max}}=\frac{343\,\frac{\text m}{\text s}}{20\,\times\,10^3\,{\text{Hz}}}$ $=0.017\,$m $=1.7\,$cm
      '
    • $\lambda_2=\frac{v}{f_2}=\frac{343\,\frac{\text m}{\text s}}{15.5\,\times\,10^3\,{\text{Hz}}}$ $=0.022\,$m $=2.2\,$cm
      '
    • $\lambda_1=\frac{v}{f_1}=\frac{343\,\frac{\text m}{\text s}}{35\,{\text{Hz}}}$ $=9.8\,$m
      '
    • $\lambda_{f{\text{min}}}=\frac{v}{f_\text{min}}=\frac{343\,\frac{\text m}{\text s}}{20\,{\text{Hz}}}$ $=17\,$m
      '
  4. Make sense out of the answer:
    • This is an impressive wavelength range from about an inch to 30-50 feet!

Example 2.2

Estimate the frequency of the “EE” sound.


Example 2.3

If one can't (almost) hear an airplane 35,000 ft above, how close can one stand next to this source of noise without feeling pain?

Assuming the power is constant, use the ratio of the threshold of pain to threshold of hearing to solve for the distance of the threshold of pain. The distance for the threshold of hearing is given. (rh = 35,000 ft = 10,668 m)

Steps:

  1. General formula for intensity:
    • $I = \frac{P}{A}=\frac{P}{4\pi r^2}$
    • $I_0=\frac{P}{4\pi r_\text{h}^2}$ $=10^{-12}\frac{\text W}{ {\text m}^2}$
    • $I_\text{t.p.}=\frac{P}{4\pi r_\text{p}^2}$ $=1\frac{\text W}{ {\text m}^2}$
  2. The ratio of these equations lets us find the distance rp at which the threshold of pain takes place (assuming the point source):
    • $\frac{I_\text{t.p.}}{I_0} =\frac{1\,\frac{\text W}{ {\text m}^2}}{10^{-12}\frac{\text W}{ {\text m}^2}}$ $= 10^{12} = \frac{r_h^2}{r_p^2}$
  3. Taking the square root of both sides:
    • $\sqrt{10^{12}}=10^6$ $= \frac{r_\text{h}}{r_\text{p}}$
    • $r_\text{p} = \frac{10668\,{\text m}}{10^{6}}$ = 0.0106 m = 1.06 cm
  4. Make sense of the answer:
    • The actual aircraft flying at 35,000 ft is actually quite audible in an otherwise quiet environment, so the actual distance for the threshold of pain will be quite a bit larger than 1 cm.

Problem 2.4

A mountaineer, whose mass (including gear) is 250 lb, is hanging next to the face of a cliff in total fog, suspended by a 24-lb, 600-ft kernmantle rope, tied to his buddy above. What is the delay in perception of the buddy's movements (hammering nails into the rock) sensed by the mountaineer through the rope?

Steps:

  1. Convert units to SI:
    • $m_m$ = 250 lb = 113 kg
    • $m_r$ = 24 lb = 10.9 kg
    • $L$ = 600 ft = 183 m
  2. Find average tension in the rope
    • $F_T=\left(m_m+\frac{1}{2}m_r\right)g$ $=118\,{\text{kg}}\cdot 9.8\frac{\text m}{ {\text s}^2}$ $=1160\,$N
  3. Find the linear mass density of the rope
    • $\mu=\frac{m_r}{L}=0.060\,\frac{\text{kg}}{\text m}$
  4. Find the speed of transverse waves in the rope
    • $v=\sqrt{\frac{F_T}{\mu}}$ $=\sqrt{\frac{1160\,{\text N}}{0.060\,\frac{\text{kg}}{\text m}}}$ $=140\,\frac{\text m}{\text s}$
  5. Find the time delay over the length of the rope
    • $\Delta t=\frac{L}{v}$ $=\frac{183\,{\text m}}{140\,\frac{\text m}{\text s}}$ $=1.3\,$s

Questions about Pr.2.4

  • Why is only half the mass of the rope considered to be contributing to the force of tension?
    • In this case, the tension force depends on the position along the rope: at the top, the tension has to support both the rope and the mountaineer, and at the bottom, it has to support just the mountaineer. Therefore, the velocity of the pulse also depends on the position: it goes a little faster as it approaches the top. An exact calculation needs to take this into account, and can become somewhat complicated $\big(\Delta t=\int_0^L{\frac{dx}{\sqrt{g\frac{m_m+\mu x}{\mu}}}}$ $=2\sqrt{\frac{L}{g}\frac{m_m}{m_r}}\left(\sqrt{1+\frac{m_r}{m_m}}-1\right)$ $=1.3112\,{\text s}\big)$. Instead, we make an approximation by taking the “average” tension, by using an arithmetic mean of the tension forces at each end of the rope ($m_mg$ at the bottom end and $(m_m+m_r)g$ at the top end). This average is $(m_m+\frac{1}{2}m_r)g$. The answer it gives is $\frac{L}{\sqrt{g(m_m+m_r/2)/(m_r/L)}}$ $=1.3109\,$s, which is pretty close to the exact answer. — Nicholas Kuzma 2014/04/21 09:55

Example 2.5

A string on a violin, 60 cm in length, is tuned to the middle A ($f_1 = 440\,$Hz) by applying 14 lb of tension. Find the mass of the string.

Steps:

  1. Convert units to SI:
    • $L=60\,{\text{cm}}=0.60\,{\text m}$
    • $F_T$ $=14\,{\text{lb}}\cdot 0.454\frac{\text{kg}}{\text{lb}}\cdot 9.8\frac{\text m}{\,{\text s}^2}$ $=62.3\,$N
  2. Fundamental harmonic $f_1=440\,$Hz
  3. Find the speed of the wave
    • $v=\sqrt{\frac{F_T}{\mu}}$ $=\lambda\,f_1=2L\,f_1$ $=1.2\,{\text m}\cdot 440\,{\text s}^{-1}$ $=528\,\frac{\text m}{\text s}$
  4. Find the linear mass density
    • $ \frac{F_T}{\mu}=v^2$
    • $\mu=\frac{F_T}{v^2}$ $=\frac{62.3\,{\text N}}{(528\,\frac{\text m}{\text s})^2}$ $=0.00022\,\frac{\text{kg}}{\text m}$
  5. Find the mass
    • $m=\mu L$ $= 0.00022\,\frac{\text{kg}}{\text m}\cdot 0.60\,{\text m}$ $=0.00013\,{\text{kg}}$ $=0.13\,$g
  6. Try to make sense of the answer:
    • Let's find the cross-section A and diameter d, assuming the string is made out of steel ($\rho_s=7800\,\frac{\text{kg}}{\,{\text m}^3}$) or plastic ($\rho_p=1000\,\frac{\text{kg}}{\,{\text m}^3}$)
      • $m=V\rho=AL\,\rho=\pi\big(\!\frac{d}{2}\!\big)^2L\,\rho$
      • $d_s=2\sqrt{\frac{m}{\pi L\rho_s}}$ $=0.19\,$mm
      • $d_p=2\sqrt{\frac{m}{\pi L\rho_p}}$ $=0.53\,$mm
    • Violin strings are pretty thin!

Example 2.6

Organ pipe (one end closed, one open) is tuned to $f_1=440\,$Hz at 20$\,^\circ$C

  1. Find its length
  2. How much does it detune at 0$\,^\circ$C?

Steps:

  1. Look up the speed of sound at these temperatures
    • $v_{20^\circ{\text C}}$ $=343\,\frac{\text m}{\text s}$
    • $v_{0^\circ{\text C}}$ $=331\,\frac{\text m}{\text s}$
  2. Write down the resonance condition for the system with opposite-type ends:
    • $f_n=\frac{nv}{4L}$, where $n=1,3,5...$
    • $f_1=\frac{v}{4L}$
  3. Find the length of the pipe
    • $L=\frac{v_{0^\circ{\text C}}}{4f_1}=\frac{343\,\frac{\text m}{\text s}}{4\,\cdot\,440\,{\text{Hz}}}$ $=0.195\,$m
  4. Find the new fundamental at the lower temperature
    • $f_1( 0^\circ{\text C} )$ $=\frac{v_{0^\circ{\text C}}}{4L}$ $=\frac{v_{0^\circ{\text C}}}{v_{20^\circ{\text C}}}\cdot 440\,{\text{Hz}}$ $=\frac{331}{343}\cdot 440\,$Hz $\approx425\,$Hz

Example 2.7

You are listening to an echo while playing a 440-Hz note and biking at 20 mph towards a wall. Find the frequency of the echo.

Steps:

  1. Convert units to SI
    • You are the source in the first Doppler effect (transmission to the wall) and, at the same time, you are the observer in the second Doppler effect (receiving the reflection):
    • $v_{s1}=v_{o2}=8.94\,\frac{\text m}{\text s}$
  2. Calculate the frequency of the sound hitting the wall (first Doppler effect)
    • the source is moving towards the observer (the wall), the observer (wall) doesn't move
    • the frequency should increase (the source and the observer are moving towards each other)
    • $f_1=\big(\!\frac{1}{1-\frac{v_{s1}}{v}}\!\big)\,f$ $=451.8\,$Hz
  3. Calculate the frequency of the sound perceived by you (the second Doppler effect)
    • the source (the wall this time) is stationary, the observer (you) is moving towards the source
    • the frequency should increase again! (the source and the observer are moving towards each other)
    • $f_2=\big(1+\frac{v_{o2}}{v}\!\big)\,f_1$ $=\left(\frac{1+\frac{v_{o2}}{v}}{1-\frac{v_{s1}}{v}}\right)\,f$ $=463.5\,$Hz
  4. Try to make sense of the answer:
    • You have to bike pretty fast to “shift” the frequency by almost half a tone (next note, 1/2 tone up, is A# = 466 Hz)
    • It also helps if you are facing forward while trying this out (please don't!)

Example 2.8

The peak blood velocity in fetal aorta is about $20\,\frac{\text{cm}}{\text s}$. If one images it using a 9 MHz ultrasound, find the maximum Doppler frequency shift of the echo at the detector. Assume the speed of sound in the body v = 1500 m/s. (The blue and red colors in the video represent the blood flow in the ultrasound image moving either towards or away from the detector. The long stripes of alternating white and gray blobs are the baby's spine, and/or the ribs penetrating the imaging plane.)

Steps:

  1. Convert all units to SI:
    • $v_{o1}=v_{s2}=0.20\,\frac{\text m}{\text s}$
  2. As with any echo problem, there are two Doppler effects:
    • First, the ultrasound source is stationary, and the “observer” is the moving blood in the aorta
      • Let's assume the blood is moving towards the ultrasound device
      • $f_1=\big(1+\frac{v_{o1}}{v}\!\big)\,f$
    • Second, the source is the moving blood, and the observer is the stationary ultrasound detector
      • $f_2=\big(\!\frac{1}{1-\frac{v_{s2}}{v}}\!\big)\,f_1$ $=\left(\frac{1+\frac{v_{o1}}{v}}{1-\frac{v_{s2}}{v}}\right)\,f$ $\approx\big(1+2\frac{v_\text{blood}}{v}\!\big)\cdot f$ $=\big(1+2\frac{0.2\,\frac{\text m}{\text s}}{1500\,\frac{\text m}{\text s}}\!\big)\cdot\,$9,000,000 Hz = 9,002,400 Hz.
  3. Find the frequency shift
    • $\Delta f=\,f_2-f=2.4\,$kHz

Homework Questions Ch.14


Problems 23(5) and 25

  • The solutions to the homework offer two pieces of conflicting information. 14:23(5) says that in a harmonic wave function, if the t and x values have opposite terms, the wave travels to the right, but in 14:25(strategy) it says that if they have opposite values, the wave travels to the left.
    • The argument of the sin or cos function, what we call $\theta$, is constant along the wave. [For example, in case of sin, when $\theta=\frac{\pi}{2}$ we have a crest.] So if $\theta$ contains $\omega t\!-\!\frac{2\pi}{\lambda}\!x\;$ or $\;\frac{2\pi}{\lambda}\!x\!-\!\omega t$ , this will be constant if both x and t increase (or decrease) - that is the wave is traveling in the $+x$ direction as time increases (so if we draw the x axis from left to right, it is to the right). If $\theta=\omega t\!+\!\frac{2\pi}{\lambda}\!x\!+\!\phi$, then it can remain constant only if x decreases with time, so the wave travels in the $-x$ direction. There must be a typo in the strategy section of solutions to Problem 25, but the answers there are correct. — Nicholas Kuzma 2014/04/07 14:05

chapter_14.txt · Last modified: 2014/05/11 18:03 by wikimanager