Sound waves You can think of a sound wave as an oscillating pattern of compression and Expansion (  P), or as an oscillating position for small packets of Air [s(x,t), which leads to the above picture]. REMEMBER this is a longitudinal wave!  = ½  2 s m 2 v s = ½(  p m ) 2 /v s   p m =  s m v s

Chapter 17 Problems NOTE:  =1.21 kg/m 3 v=343 m/s At T=20 o C FIRST: What is the ratio of the Intensities between the two cases? e). What is the pressure difference that corresponds to each of these intensities? Finish off this to start off on Monday.

Interference with Sound waves Suppose the difference L2-L1 is 1.70 m and S1 and S2 emit sound at A frequency of 100Hz. How will the intensity at P change if the phase Offset between S2 and S1 is changed from zero to  ? (assume v=340m/s)

Chapter 17 Problems

Reflections at a Boundary

Standing waves: Pipes The resonance frequencies of pipes depend on the conditions at the two ends. A closed end needs a NODE, and an open end needs an ANTINODE. The book gives you formulae for the two cases that you can remember, OR you can just remember these two conditions and draw pictures! (I find this way of doing it MUCH easier.) Question: Why do we not consider the case of both ends closed? What would be the condition in that case?

Chapter 17 Problems

If the amplitude of a sound wave is doubled, by what number of dB does the intensity of that sound wave increase? As usual, please provide a brief explanation for your answer (16 did not answer) if the amplitude of a sound wave is doubled then the intensity of the wave is quadroupled and the sound wave increases by 40dB (8 or so made various mistakes such as this) since intensity is directly related to the square of the amplitude, when the amplitude is doubled the intensity would quadruple. (9 answered this way) If the amplitude of a wave was doubled, its intensity would be multiplied by 4. This would translate to an increase of approximately 6 dB. (8 answered like this)

Standing waves: Pipes The resonance frequencies of pipes depend on the conditions at the two ends. A closed end needs a NODE, and an open end needs an ANTINODE. The book gives you formulae for the two cases that you can remember, OR you can just remember these two conditions and draw pictures! (I find this way of doing it MUCH easier.)

Chapter 17 Problems

Doppler Effect The frequency shifts up if the source and observer are getting closer to each other, and down if they are receding from each other. Think of old “murder on the train movies”, or the sound of an Indy car as it goes by. NOTE: this is the phenomenon that is used to measure the changes in the velocity of stars so accurately that extra-solar planets can be detected (e.g. problem 13-52); it works for light and all kinds of waves, not just sound!

Two identical loud speakers are emitting sounds at a frequency of 130 Hz, but one of the two is on the ground and the other is on a flat bed rail car moving at a speed of 10 m/s what beat frequency is heard by an observer on the ground who views the rail car approaching him? (take the velocity of sound to be 343 m/s). Please give a brief explanation of how you got your answer. (17 no answers, 5 confused) f'=f[v/(v+v_s)] = 126 Hz (source moving, detector stationary) (careful, ask yourself if the frequency would be greater or smaller as a result of the motion; also this does not address the beat freq. question; 6 went this route). f=(130Hz)((343m/s+10m/s)/(343m/s))=133.8Hz f(beat)=133.8Hz-130Hz=3.8Hz ( 10 got 4 Hz, and another 5 got the 134 Hz for the result of the Doppler shift, but forgot to compute the beat freq. BUT NOTE: some, like this one got lucky; the source is moving so the 10m/s should be in the denominator with a minus sign not in the numerator with a plus sign!)

BEATS If the frequencies are not matched, then the interference changes from constructive to destructive periodically in time as the higher- frequency wave picks up an extra  phase shift. The “Beats” show a maximum every time there is a 2  phase shift (i.e. the higher frequency wave picks up a whole cycle on the lower frequency wave). Beat frequency is just the difference in the frequencies of the two waves.

Chapter 17 Problems

Extra Office Hours Exam Week Monday (28 April): –DVB: 1:30 to 3:30 –Tayloe: 2:00 to 4:00 Tuesday (29 April): – DVB: 11:00 to 12:00 and 1:30 to 3:30 –Tayloe: 9:00 to 11:00. Final Exam is at 8:00-10:00 on Wednesday 30 April 2008 in SW 007

Final exam Will be comprehensive! ~1/4 - 1/3 of the questions will be on the new stuff (since exam III) Will be out of approximately 120 points (i.e. about 50% longer than the midterms but you have more than twice as long). ~6 multiple choice 4 or 5 multi-part questions.

Review requests for Final New stuff/oscillations/waves (13 requests) Ang. momentum/torque etc. (4 requests) Fluids (2 requests) All else, no more than 1 request

What is the fundamental definition of temperature? The measure of thermal energy on a ˜body˜. Temperature is the measure of the average kinetic energy of a system of particles. (~9 something like these) When two bodies are in contact the ˜colder˜ body absorbs heat from the ˜Warmer˜ body. Temoperature is a measurement which describes the ˜warmness˜ of each body. (15 like this)

1.What physical phenomenon is used to determine the temperature in a common alcohol (or mercury) thermometer. When temperature increases and volume remains constant, pressure increases. An increase of pressure inside a thermometer makes the alcohol rise to maintain equilibrium. (5 like this) The triple point of water is used to determine the temperature in a common thermometer. This is the point where the pressure and temperature are just right so that liquid water, ice, and water vapor all coexist. (3 like this) Thermal expansion: As an object heats up the object expands. (9 like this; linear expansion or volume?)

Chapter 18 Problems NOTE: the cross sectional area changes! In Thermal expansion, all linear dimensions change (length of edges, sides of holes etc.) by the same fraction for a given  T

Thanks for a great semester!!