State Oregon University
PH 202 Midterm Exam 2 (Wednesday, February 15, 7:00 p.m.)
Winter Term 2012
GENERAL DIRECTIONS: Fill out the cover sheet completely, as indicated--including your test form letter. Follow the specific directions on each page of the answer form. Use standard SI units unless directed otherwise; use three significant digits in all final numerical answers. Questions #1-4 are worth 12 points each; #5-7 are worth 15 points each; #8-9 are worth 18 points; #10 is worth 21 points.
Physical constants and other possibly useful information (such as E, your test form letter).
1. Refer to the diagram below. At (A) IB = 2IA. point X, which statement is true? (B) bB > (bD + 10 dB) (C) Sound from C will be twice as loud to your ears (brain) as sound from A. (D) All of the above. (E) A and B only. (F) A and C only. (G) B and C only. (H) None of the above. 2. In the diagram below, assume for (A) If d = (7.5)l, then B and C produce Destructive Interference (D.I.). this question that all speakers are (B) If d = (4.0)l, then B and D produce Destructive Interference (D.I.). motionless and that only the two (C) If d = (0.5)l, then A and D produce Constructive Interference (C.I.). mentioned in each statement are (D) All of the above. sounding. At point X, which (E) A and B only. statement is true? (F) A and C only. (G) B and C only. (H) None of the above. 3. Refer to the diagram below. At (A) Sound from A is louder than sound from C. point X, which statement is false? (B) You hear a higher frequency from B than from D. (C) Speaker C is more powerful than speaker D. (D) All of the above. (E) A and B only. (F) A and C only. (G) B and C only. (H) None of the above. 4. Which statement is false? (A) If you send the the same frequency of sound through 10C air and through 30C air, the wavelengths will be longer in the warmer air. (B) If a 1-kg mass were oscillating horizontally on a spring of stiffness 4 N/m, a pendulum of arm length 2.45 m would have about the same frequency. (C) Simple harmonic motion can be described by either the x- or y-motion of an object in uniform circular motion. (D) All of the above. (E) A and B only. (F) A and C only. (G) B and C only. (H) None of the above.
g = 9.80 m/s2
1 atm = 1.01 x 105 Pa
I0 = 10-12 W/m2
Speaker A
d
Speaker B
d
Point X
|
d
Speaker C
3d
Speaker D
Four identical speakers, with current positions as shown, are emittting identical (spherical) sound waves (same power, one frequency, in phase). The speakers are all moving toward point X, and vA = vB = vD > vC.
5. Two guitar strings (each fixed at both ends) with identical lengths and identical linear mass densities are vibrating at their fundamental frequencies. String A is vibrating at 180 Hz, but because both strings vibrating, you hear a beat frequency of 5 hz. If you were to tighten (increase the tension of) string B, that would increase the beat frequency. But you're allowed to adjust only string A. By what percentage must you tighten (increase the tension of) string A to equalize the frequencies of the two strings? 6. Which statement is true? (A) If a tube of air is open at one end, closed at the other, and is vibrating at its 3rd harmonic frequency, there is a pressure node at the tube's midpoint. (B) As a sound wave travels westward through air, at any given moment in time, it may include individual particles that are east of their rest positions. (C) If you're standing still, a horn emitting a 220 Hz tone will sound 1 octave higher to you if it is moving toward you at half the speed of sound. (D) All of the above. (E) A and B only. (F) A and C only. (G) B and C only. (H) None of the above. 7. A tube, open at both ends, with a diameter of 1.64 m, is used as a foghorn on the Oregon coast. It faces directly west --out to sea (and assume the coastline runs straight north and south). Weather varies throughout the year, so the speed of sound ranges from 325 m/s to 360 m/s. The tube resonates at its fundamental frequency, with a power of 1000 W. a. What is the farthest distance from the foghorn at which a person could theoretically just (barely) hear it? b. Of course, we don't want just barely-audible sound for mariners who are trying to avoid coastal boating hazards. So, what is the farthest distance from the foghorn at which it could be heard at 60.0 dB?
c. What is the minimum length of the tube so that its sound could be heard by any boat at any location on the ocean surface (within the 60-dB distance)--at any time of year? 8. You have three identical, empty beverage bottles, A, B and C (each open at the top and closed at the bottom), each with an inside length of 25 cm. Assume the speed of sound in air is 344 m/s. Using the least water possible, add whatever water you judge necessary to any bottle (A gets the least water; C gets the most), so that when air is blown across all the bottles, their fundamental frequencies produce a major chord (i.e. frequency ratios of 4:5:6). For each bottle, what is the required water level (as measured in cm from the bottom of the bottle)? 9. Which statement is false? (A) If bA = 80 dB and bB = 85 dB, then IB/IA < 3.50. (B) If two mowers (same power and distance away) are operating, then one shuts off, the sound to your ears (and brain) is then half as loud as before. (C) For a guitar string (fixed at both ends, with constant length, tension and linear mass density), its 3rd harmonic frequency is 2 octaves higher in pitch than its fundamental frequency. (D) All of the above. (E) A and B only. (F) A and C only. (G) B and C only. (H) None of the above. 10. Two identical loundspeakers are emitting (in phase) the same sound frequency, f. On x-y coordinate axes, speaker A is located at the point (dA, 0); and speaker B is located at the point (dB, 0). Sound from speaker A arrives at the origin exactly two full cycles sooner than sound from speaker B. Sound from speaker B requires a time Dt to travel all the way to speaker A. Starting from the origin... a. What non-zero distance along the positive y-axis must you move to arrive at the nearest point of C.I.? b. What non-zero distance along the negative y-axis must you move to arrive at the nearest point of D.I.?
This is a "No-Solve" problem. There are no numbers provided for this problem; you are not being asked to get a numeric solution or to do any math at all--not even any algebra. Rather, you are to write a series of succinct instructions on how to solve the problem. Pretend that your instructions will be given to someone who knows math but not physics. Again, don't do any algebra--no re-arranging or anything. Just tell another person how to do it.
E
Consider these values as known (and all of these are positive values): f, dA, dB, Dt.
Reminder: Before turning in your exam, be sure to follow the directions on the front page here. Fill out the cover sheet accordingly.
LAST 4 digits of OSU ID
For each question to be answered on this page... LEGIBLY PRINT your ONE answer in the smaller (dotted-line) space: For a multiple choice question, write a single letter answer. For a numeric answer, use 3 significant digits and proper units. For a word answer, be brief and specific--no multiple answers. Partial credit may be awarded, but only for work shown in the larger space.
LAST name (first 4 letters)
1. Answer: B A C D E F G H 2. Answer: C A B D E F G H 3. Answer: D A B C E F G H 4. Answer: H A B C D E F G
Oregon State University
(12 points) (4 points) (4 points) (4 points) (8 points) (0 points) (8 points) (8 points) (12 points) (4 points) (4 points) (4 points) (0 points) (8 points) (8 points) (8 points) (12 points) (4 points) (4 points) (4 points) (8 points) (8 points) (8 points) (0 points) (12 points) (8 points) (8 points) (8 points) (0 points) (4 points) (4 points) (4 points)
PH 202, Winter Term 2012 Midterm Exam 2 (Wed., February 15, 7:00 p.m.) Page 3
LAST 4 digits of OSU ID
For each question to answered be on this page... LEGIBLY PRINT your ONE answer in the smaller (dotted-line) space: For a multiple choice question, write a single letter answer. For a numeric answer, use 3 significant digits and proper units. For a word answer, be brief and specific--no multiple answers. Partial credit may be awarded, but only for work shown in the larger space.
LAST name (first 4 letters)
5. Answer: 5.63%
15 pts total: units, sf's = 1/2 pt. ea. 2 pts. 2 pts. 3 pts.
Which frequency is higher? When you increase the tension of string B (which increases fB), that increases fbeat.; you're increasing the difference between the two pitches. So fB is greater than fA already. Thus: fbeat = fB fA = 5 So: fB = 5 + fA = 5 + 180 = 185 Hz. So, to tune the two strings, you must adjust fA upward, from 180 Hz to 185 Hz. That is, for string A: ff / fi = 185/180
The beat frequency is the difference between the two strings' frequencies: fbeat = | fA fB |
The fundamental frequency of a string (fixed at both ends) is given by f1 = vstring./(2L) Thus: [vstring.f/(2L)]/[vstring.i/(2L)] = 185/180 Or: vstring.f/vstring.i = 185/180 The speed of a wave on a string (fixed at both ends) is given by vstring = (FT/m) Thus: [(FT.f/m)]/[(FT.i/m)] = 185/180 Or: FT.f / FT.i = (185/180)2 And:
3 pts. 3 pts. 2 pts.
D%FT = [(FT.f FT.i)/FT.i](100) = [(FT.f / FT.i) 1](100) = [(185/180)2 1](100)
6. Answer: G A B C D E F H
(15 points) (0 points) (10 points) (10 points) (10 points) (5 points) (5 points) (5 points)
7. Answers: (see below)
So: I0 = P/(4prmax2) So: I60 = P/(4pr602)
15 points total: units & sig figs. = 1/2 pt. ea.
a. A sound barely audible to a person has an intensity of I0 = 10-12 W/m2. b. A 60-dB sound has an intensity given by: 60 = 10log(I60/I0)
Thus: rmax = [P/(4pI0)] = [1000/(4p10-12)] = 8.92 x 106 m Thus: r60 = [P/(4pI60)] = [1000/(4p10-6)] = 8.92 x 103 m (8.92 km) Solving: I60 = 10-6 W/m2.
3 pts. 2 pts. 3 pts. 2 pts. 2 pts. 3 pts.
c. To be heard all over the (local) ocean surface, the horn must have a dispersion angle q = 90: The horn's aperture is circular: sin90 = 1.22l/D That is: 1.22l/D = 1 or: l = D/1.22 But for the tube's fundamental harmonic mode: L = l/2 or: l = 2L Substitute: 2L = D/1.22 Thus: L = D/(21.22) = 1.64/2.44 = 0.672 m
Oregon State University
PH 202, Winter Term 2012
Midterm Exam 2 (Wed., February 15, 7:00 p.m.)
Page 4
LAST 4 digits of OSU ID
For each question to be answered on this page... LEGIBLY PRINT your ONE answer in the smaller (dotted-line) space: For a multiple choice question, write a single letter answer. For a numeric answer, use 3 significant digits and proper units. For a word answer, be brief and specific--no multiple answers. Partial credit may be awarded, but only for work shown in the larger space.
LAST name (first 4 letters)
8. Answers: (see below)
In an air-filled tube of length L, closed at one end but open at the other, each harmonic frequency must have a wavelength that satisfies L = n(ln/4), where n is an odd integer. Re-expressed in terms of frequency this becomes fn = nv/(4L), where v is the speed of sound in air. Here L is different for each bottle (its length of vibrating air shortened by water), but n = 1 always, since each must vibrate at its own unique fundamental frequency. Thus, for each bottle: f = v/(4L) We want these ratios among those frequencies: fB/fA = 5/4 and fC/fA = 6/4 (or fC/fB = 6/5) a. The greatest L produces the lowest frequency; every higher frequency will require a shorter L --more water. So, to minimize the water used, leave bottle A empty (thus LA = 25 cm): The water level in bottle A is zero (0.00 cm). And so bottle A's fundamental frequency will serve as the lowest note in the chord: fA = v/(4LA) = 344/(40.25) = 344 Hz. b. fB/fA = 5/4. That is: [v/(4LB)]/[v/(4LA)] = 5/4 Simplify: LA/LB = 5/4 Solving: LB = (4/5)LA = (4/5)(25) = 20 cm. That's 5 cm shorter than the length of air vibrating in an empty bottle. The water level in bottle B is 5.00 cm. c. fC/fA = 6/4. That is: [v/(4LC)]/[v/(4LA)] = 6/4 Simplify: LA/LC = 6/4 Solving: LC = (4/6)LA = (4/6)(25) = 16.67 cm. That's 8.33 cm shorter than the length of air vibrating in an empty bottle. The water level in bottle C is 8.33 cm.
2 pts. 2 pts. 2 pts. 2 pts.
2 pts.
4 pts.
4 pts.
9. Answer: G A B C D E F H
(18 points) (0 points) (12 points) (12 points) (12 points) (6 points) (6 points) (6 points)
Oregon State University
PH 202, Winter Term 2012
Midterm Exam 2 (Wed., February 15, 7:00 p.m.)
Page 5
LAST 4 digits of OSU ID
For problem #10, use the front and back sides here, as needed. And note: If the question requires a numeric answer, use 3 significant digits and proper units in your answer. (Partial credit will be awarded, so show all work. No credit will be awarded for an answer without valid work shown.) If this is a "no-solve" question not requiring a numeric answer, do no math --no algebra, no "re-arranging" or simplifying or anything. Simply write a series of succinct instructions on how to solve the problem. Pretend your instructions will be given to someone who knows math but not physics.
LAST name (first 4 letters)
10. Two identical loundspeakers are emitting (in phase) the same sound frequency, f. On x-y coordinate axes, speaker A is located at the point (dA, 0); and speaker B is located at the point (dB, 0). Sound from speaker A arrives at the origin exactly two full cycles sooner than sound from speaker B. Sound from speaker B requires a time Dt to travel all the way to speaker A. Starting from the origin... a. What non-zero distance along the positive y-axis must you move to arrive at the nearest point of C.I.? b. What non-zero distance along the negative y-axis must you move to arrive at the nearest point of D.I.?
This is a "No-Solve" problem. There are no numbers provided for this problem; you are not being asked to get a numeric solution or to do any math at all--not even any algebra. Rather, you are to write a series of succinct instructions on how to solve the problem. Pretend that your instructions will be given to someone who knows math but not physics. Again, don't do any algebra--no re-arranging or anything. Just tell another person how to do it.
Consider these values as known (and all of these are positive values): f, dA, dB, Dt.
I. Find the speed of sound: Use this equation: vsound = (dA + dB)/Dt Use these knowns: dA (given) dB (given) Dt (given) II. Find the wavelength of the sound being emitted by the speakers: Use this equation: vsound = fl Use these knowns: vsound (from part I) f (given)
Solve for: vsound
3 pts.
Solve for: l
3 pts.
III. It takes two extra cycles for sound B to arrive at the origin (and the time for each cycle is the period, T). So sound B has to travel two extra wavelengths of distance: Use this equation: dB dA = 2l (Note: Solving for l directly here earns all above points, too.) IV. From the result of part III, we see that the origin is a point of Constructive Interference (C.I.); the path length difference is a whole multiple (2) of the wavelength. Use this equation: Dx = nl Use these knowns: Dx = dB dA Solve for: n (n = 2) V. Now, as we move away from the origin along the y-axis, the path length difference, Dx, will decrease. (To see this, use some appropriate example values for dB, dA and l, then choose a new location on the +y-axis and calculate Dx there. It will be less than 2l.) VI. So the nearest point of C.I. on the positive y-axis--call it (0, yCI)--will be where n = 1: Use this equation: Dx = (1)l Where: Dx = (dB2 + yCI2) (dA2 + yCI2) Solve for: yCI l (from part II) VII. And the nearest point of D.I. on the negative y-axis--call it (0, yDI)--will be where n = 1.5: Use this equation: Dx = (1.5)l Where: Dx = (dB2 + yDI2) (dA2 + yDI2) Solve for: yDI
Oregon State University PH 202, Winter Term 2012 Midterm Exam 2 (Wed., February 15, 7:00 p.m.)
3 pts.
3 pts.
3 pts.
3 pts.
3 pts. Page 6