PHY2464 Fall 2005 Sample Exam with solutions – Hall, Chapters 1 – 5

For this and similar exams you are permitted to bring one standard-size [8 ½” x 11”] page of notes and a calculator. Any constants [not formulae!] you need that are not included with the exam will be provided upon request.

Useful (?) Constants

v= 344 m/s at 20 C g = 9.8 m/s² area of sphere = 4pi r²

Here are some typical questions of the type that might be asked, in addition to the homework problems/questions. Some might be in multiple-choice format. [Since these take longer to compose, few are included here]. ‘SW’ following a question number means “Show Work”: all calculations must be clearly written out. Answers without calculations will receive little credit, even if correct.

In composing the solutions, some questions have been edited for clarity. Note that this browser  doesn't reproduce all symbols or Greek letters so some [lambda, pi] have been spelled out.

1 SW. A foghorn’s sound intensity is measured to be 1 W/m² at 500 m .

What is the intensity at 1 km?

[See Section 5.3] I is proportional to 1/ r² ; r1 = 0.5 km and r2 = 1km = 2 r1 . So:

I1/I2 = r₂ ² / r₁ ² . So I2 = I1 [ r₁ ² / r₂ ² = 1 W/m² [1km/2km]² = 1/4 W/m² or 0.25 W/m²

2 SW. Please refer to the above question. What is the power output of the foghorn?

The sound waves leave the source and (unless intercepted by various objects) spread spherically. Thus at 500 m, the source power has spread over a sphere of radius 500 m.

So, I = P/4pi r² => 1 W/m² = P/ 4 pi (500 m)²

P = 4 pi (25 0000 m²) 1 W/m² = 3.14 x 10⁶ W or 3140 kW [this is probably an impossibly powerful foghorn!]

3 SW. Please see question 1, above. What is the sound intensity level [SIL] at 500m?

[See 5.2] SIL = 10 log (I/Io) = 10 log 1/10 ^-12 = 10 log (10¹²) = 10 (12) = 120 dB

4 SW. An organ pipe originally tuned to A = 440 Hz at 20 C produces 5 beats per second when compared with a standard tuning fork. What is the room temperature.

assuming that the pitch change is due to temperature?

There are 2 possibilities: T may be either above or below 20C making the pitch either sharp or flat. If T is higher, 5 beats/s => the pipe f = 445 Hz.

Since v = f l, f µ v [the wavelength doesn’t change appreciably] , so [see 1.4],

5 = 0.6 (T -20C) = 0.6T - 12 => T = (5 +12)/ 0.6 = 28.3C

[So, T increased by 8.3C. Try working out the solution for decreased T]

5. The speed of sound in air at 10 C will be (in m/s):

A.  338 B. 350 C. 344 D. 330 E. none of these

[See 1.4] . Verify for yourself that answer A [v= 338m/s] is correct.

6. How many violins would be required to sound four times as loud as one violin?

Assume the single violin plays at SIL = 60 dB.

Assuming that “four times as loud” means “4 times the intensity”, 4 violins. Note that this is not the same as “four times the intensity level”!!

7. The note range on a piano is approximately 32 Hz – 4186 Hz. To what approximate

wavelengths do these frequencies correspond?

[See 2.1] Assuming 20C, 32Hz corresponds to lambda = 344m/s/32 Hz = 10.75 m and

4186 Hz to 344/4186 = 0.082m or 8.2 cm

8. You are standing on the field, directing the Gator Marching Band. Two friends of yours, both playing trumpet, march toward you from opposite directions as they play treble C (approximately 500 Hz). If each marches at 1 m/s, will they sound out of tune to you? Support your answer with a calculation.

This question, as stated, is ambiguous: does ‘sound out of tune’ mean that you have a way of deciding their individual pitches, or does it mean ‘with respect to each other’??

(a) If the latter, the answer is ‘NO’: Even if the players are moving fast enough to produce detectable Doppler shift [see next answer], they are moving toward you at equal speeds from opposite directions so you’ll hear the same pitches/frequencies as if they were stationary.

(b) The Doppler-shifted frequency [see 4.4] for each moving player is :

f = fo/(1-V/v) = 500Hz/ [1- 1m/s/344m/s) = 500Hz/ (1 – 0.0029) = 500Hz/0.9971 =

501.45Hz. So if you(the “observer”) had perfect pitch or a standard sound source, the shifted frequency would produce 1.45 beats/s which you could presumably hear.

9. Take the distance between your ears to be 20 cm. At what frequencies (if any)

might you expect to have difficulty in distinguishing the location of sound

sources, assuming that you are blindfolded? Justify your reasoning.

[See 4.2] This is analogous to problem 4.2: sound waves with wavelengths appreciably longer than 20 cm will be diffracted by your head. The diffracted waves will be difficult to “place’.

What’s ‘appreciably longer’?? Pick a convenient number: 1m is 5x longer and will be easy to work with.  Thus f = (344m/s)/1m = 344 Hz.  So sound sources producing frequencies below around 300Hz will be increasingly difficult to locate, although moving your head might help [why?]. Perhaps you’ve noticed that deep bass notes from a pipe organ seem to “fill the room” which means that they seem to have no source location.

10. Before the development of radar, distant approaching aircraft were detected by

two large movable horns. The engine/propeller sounds picked up by each horn

were amplified and transmitted separately to each ear of the observer. Would this

work? Explain.

Yes, it works.  Interference  [see 4.5] will produce appreciable differences in what each ear hears, depending on how the horns are aimed:

H<

O………………………… ,,,,,,,,,,,,,,,A

H<

The large [1 m or more across the open ends] horns effectively give the observer bigger ‘ears’ that receive more of the incoming waves’ intensity. If the distant aircraft A is moving along the center line between the horns H< ,  the observer O will hear the combined frequencies of the amplified sound waves. The waves will be in phase because they are traveling equal distances; thus O hears more sound. If the horns are slowly rotated around a point midway between them, one horn moves closer to A while the other moves away. Waves from A reaching the two horns will go out of phase and the observer will hear less sound, so by rotating the horns and finding the direction of maximum sound intensity the general location of the aircraft can be found.