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Physics 203 College Physics I Fall 2012

Physics 203 College Physics I Fall 2012. S. A. Yost. Chapters 11 – 12 . Waves and Sound. Announcements. We will discuss Waves and Sound today: Sec. 7 – 9 in Chapter 11 and sec. 1 – 4 and 7 in Chapter 12. H omework set HW12 on this is due Thursday.

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Physics 203 College Physics I Fall 2012

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  1. Physics 203College Physics IFall 2012 S. A. Yost Chapters 11 – 12 Waves and Sound

  2. Announcements • We will discuss Waves and Sound today: Sec. 7 – 9 in Chapter 11 and sec. 1 – 4 and 7 in Chapter 12. • Homework set HW12 on thisis due Thursday. • It is relatively short, taking into account that it is due in 2 days. You will need to read some topics before we discuss them in class. • The Final Exam is next Monday: 8 – 11 AM here. • Bring a page of notes. I won’t provide equations. I will provide constants, conversion factors, and moments of inertia.

  3. Clicker Question • Are you here? • A = Yes • B = No • C = Both • D = Neither • E = Can’t be determined

  4. Wave Motion • A disturbance that moves through the medium is called a traveling wave. • A traveling wave could be continuous, or just a one-time pulse as shown here: v

  5. Wave Motion • A pulse could be created by pulling up sharply on one end of a rope. Nearby parts of the rope are pulled up a little later, and the pulse moves down the rope. v

  6. Wave Motion • Continuously moving the end up and down will make a continuous wave, or periodic wave.

  7. Wave Motion • The distance between peaks is the wavelength l. l

  8. Wave Motion • The height of the peaks is called the amplitudeA. A

  9. Wave Motion • The crests of the wave move with wave velocity v down the rope. v

  10. Wave Motion • The material of the rope doesn’t travel – it just vibrates up and down, while the wave moves through it.

  11. Wave Motion • The time for the wave to travel one wavelength is called the periodT. l

  12. Wave Motion The wavelength, period, and velocity are related: • l = v T. l

  13. Wave Motion • The frequency is the number of crests per unit time that pass a point: 2 1 5 3 4 0 seconds 8 6 2 9 10 1 7 5 3 4

  14. Wave Motion • In this case, 5 crests passed in 10 seconds, so the frequency is • f = 5/10 s = 0.5 s-1

  15. Wave Motion • The frequency is measured in units called Hertz: 1 Hz = 1 s-1 • This example had a frequency of 0.5 Hz.

  16. Wave Motion • If 5 cycles pass in 10 seconds, each cycle lasted T = 2 s. This is the period. • In general, T = 1/f.

  17. Wave Motion • The wave velocity is the product of the frequency and wavelength: • v = l/T = l f.

  18. Wave Speed • The stronger the restoring force when a displacement is made from equilibrium, the faster the wave travels. Waves move slower in a denser material, because more massive materials have more inertia, and resist propagating the wave. Wave on a string of linear mass density m = m/L under tension FT: v = √ FT / m

  19. Standing Waves • If a string is tied down at both ends, it can have a standing wave, which doesn’t go anywhere, but vibrates up and down in place with the ends fixed. • L • There has to be an integer number of half-wavelengths on the string, so l = 2L/n • The mode above, n = 1, is called the fundamental mode.

  20. Standing Waves • The next mode is n = 2. This is called the second harmonic. A full wavelength fits between the ends. • L • The relation lf = v still holds, with v determined by the tension of the string and its mass… • The mode above has twice the frequency of the fundamental.

  21. Piano String Example • A 1.6 meter steel piano wire with a mass of 38.4 g put under a tension of 1200 N. • What is the fundamental frequency heard when the string is struck?

  22. Piano String Example • The fundamental frequency can be found from the condition L = l/2whereL = 1.63 mis the length of the stretched piano string andlis the wavelength of a standing wave with nodes at the end of the string. • Thenf = v/l, wherevis the wave speed on the string. The vibrating string will cause he air to vibrate at the same frequency.

  23. Piano String Example • To find the wave speed v, we need the relation • v = √ FT/m • where FT= 1200 N is the tension on the string andm is the linear mass density of the steel spring. What is m?

  24. Piano String Example • The linear mass density of the piano wire is • m = m/L = 0.0384 kg/1.6 m = 0.024 kg/m • Since FT = 1200 N and m = 0.024 kg/m • v = √ FT/m =224 m/s. • The frequency is then • f = v/l = 224 m/s / (2L) • = (224 m/s) / (3.2 m) = 70 Hz.

  25. Sound Waves • But how do we hear the piano string? • The vibrations must move from the string into the nearby air. • Vibrations of air molecules carry the wave to our ears – we call this a sound wave.

  26. Sound Waves • Sound waves are longitudinal pressure waves where the air molecules periodically get closer together and further apart, causing variations in the air pressure.

  27. Sound Waves • v = fl is different for sound waves and the vibrating object. For sound waves v = 343 m/s at ordinary pressures and temperatures (20oC). For our piano wire, v was 224 m/s.

  28. Sound Waves • Since each vibration of the object creates one vibration of the air, the frequencies(vibrations per second) are the same when a wave passes from one media to another.

  29. Piano String Example • What is the wavelength of the sound wave produced by this piano string? • The frequency in air is also f = 70 Hz since the air vibrates together with the string. • The speed of sound in air is vs = 343 m/s. • Then the wavelength of the sound waves is • ls = vs/f = 4.9 m. Greater than the wavelength on the string

  30. Energy in a Spherical Wave • If energy carried by a spherical wave is conserved, then the power flowing through any sphere of radius R is equal to the power radiated by the source, assuming no loss. R Source All the power has to flow through this sphere unless there is absorption on the way.

  31. Energy in a Spherical Wave • Intensity is power per unit area: I = P/A. • The area of a sphere of radius R isA = 4pR2. • Energy conservation: P = 4pR2Iat any radiusR. R Source

  32. Energy in a Spherical Wave • The intensitiesI1 and I2 at distances R1and R0from the sourceare related by an inverse-square law: • I2/I1= (R1/R2)2. inverse square law for wave intensity. R1 R2 Source

  33. Loudness • Our ears can perceive a very wide range of sound intensities. • The threshold of hearing is the quietest sound a normal person can hear. • This depends on frequency, but is about • I0 = 10-12 W/m2 • (at a frequency of 1000 Hz).

  34. Loudness • I0 = 10-12 W/m2 is taken to be a standard measure of the minimum audible sound intensity. • The loudest sounds we can hear (without extreme pain and excessive ear damage) is about 1 W/m2. • This means we can hear 12 orders of magnitude (powers of 10) in variations of sound intensity! • Such a wide range of intensity is best represented by a logarithmic scale, which counts powers of 10.

  35. Decibel Scale • The sound level of an intensity I in decibels (dB) is given by • b = 10 log10 (I/I0) • (b is used for decibels in Giancoli, but is not necessarily a standard notation.) • I0= 10-12 W/m2 corresponds to 0 dB, • Imax= 1 W/m2 corresponds to 120 dB.

  36. Logarithms • Some useful properties of logarithms: • log10 (10N) = N • For any base, • log (1) = 0 • log (xy) = log x + log y • log (x/y) = log x – log y

  37. Example • How loud is a drum if the intensity is measured to beI = 3.2 x 10-2 W/m2at a distanceR = 5 mfrom the drum? • The sound level in decibels is • b = 10 log10 (I/I0) = 10 log10 (3.2 x 1010) • = 105 dB

  38. Decibel Scale • When comparing two intensities, the decibel levels are related by • b2-b1= 10 log10 (I2/I1) • This is useful when comparing any two loudness levels, without having to use the threshold of hearing.

  39. Example • How loud would the drum be at a distance of 10 m (twice as far)? • The intensity would be ¼ as great, due to the inverse square law for the intensity of spherical waves. (Power = Intensity x area = constant) • This would reduce the loudness by • Db = 10 log10(1/4) = - 6.0 dB

  40. Example • The loudness of the drum at 10 m would then be • 105 dB – 6.0 dB = 99 dB. • Note that a factor of 4 in intensity is 6 dB, and a factor of 2 in intensity is 3 dB.

  41. Wind Instruments • Wind Instruments produce sounds using a vibrating column of air. • For example, consider this tube, open on the ends.

  42. Wind Instruments • Air can vibrate back and forth, but at the ends, the pressure must be the same as it is outside the tube. • The air can blow in and out freely, keeping the pressure fixed. So there is no pressure variation at the ends.

  43. Wind Instruments • Drawing a red line for the difference between the atmospheric pressure and pressure in the pipe, the wave’s pressure maxima and minima would look like this: L antinode node node

  44. Wind Instruments • Then the frequencies produced by the open tube are • fN = Nv/2L = Nf1 • A tube that is closed on one end behaves differently. Read about that case in the book before doing the homework. L antinode node node

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