2. Sleeping Beauty has been out late with Prince Charming again and she wants to

Question

2. Sleeping Beauty has been out late with Prince Charming again and she wants to catch up on her sleep during her Music 101 class. The Music 101 professor uses a microphone in the large lecture hall. Her voice is amplified over two speakers located at the front of the lecture hall in the upper right and left corners, separated by a distance of D=15 meters. Sleeping Beauty picks out a seat that is located S1= 13.0 meters from right speaker and S2 = 12.5 meters from the left speaker. Assume that the speed of sound is 343 m/s.

a) Assuming that the speakers are connected with the same phase, what frequencies will be canceled out, or at least be minimized?

b) What frequencies will be maximized?

c) Now the Music professor connects separate frequencies to each of the speakers. The right speaker is putting out a pure tone of 440 Hz. The left speaker is putting out a lower frequency. Sleeping Beauty hears a beat frequency of 14 Hz. What is the frequency coming out of the left speaker?

d) As a demonstration of a rather unusual musical instrument, the professor swings a length, L= 0.7 m of flexible tubing over her head in a circle. The tube is open at both ends. What is the lowest frequency she can obtain?

e) The Music professor has gone back to the speakers, and has now arranged them so that the same frequency f = 6860 pure tone is coming out of each speaker. Now Sleeping Beauty decides that she cannot get any sleep here so she gets angry and exits from the front of the lecture hall. As she is leaving she passes directly between the two speakers walking directly from one speaker and directly towards the other one. She hears a beat frequency of 120 Hz . How fast is she walking?

Explanation / Answer

given D = 15 m

S1 = 13 m

S2 = 12.5 m

a. for the speakers copnnected in phase, path difference, d = S1 - S2 = 0.5 m

so the wavelengths cancelling out be lambda

then

d = (2n - 1)lambda/2

lamdba = 2d/(2n - 1) [ where n = 1,2,3, ...]

now speed of sound in air, c = 343 m/s

hence frequencies cancelling out = f

f = c/lambda = (2n -1)c/2d

f = 343(2n - 1)/2*0.5 = 343(2n - 1) Hz

so f = 343, 1029, 1715, 2401 Hz etc will be cancelled out

b. for maxima condition

f = c/lambda

lamdba = d/n

f = nc/d = 686n ( n = 1,2,3,..)

hence

f = 686, 1372, 2058, 2744 Hz will become maximum

c. frequency from speaker 1, f1 = 440 Hz

frequency from second speaker = f2

frequency heard, f = 14 Hz

path diff, d = 0.5 m

now, let the amplitude from the two speakers be same = A

then equation of the waves form two speakers is

y1 = Asin(2*pi*f1*t - 2*pi*f1*x/c)

y2 = Asin(2*pi*f2*t - 2*pi*f2*x/c)

now, x1 = S1 = 13 m

x2 = S2 = 12.5 m

then equation of the waves form two speakers is

y1 = Asin(2*pi*f1[t - 13/334])

y2 = Asin(2*pi*f1[t - 12.5/334])

y1 + y2 = A[sin(2*pi*f1[t - 13/334]) +sin(2*pi*f1[t - 12.5/334]) ]

y1 + y2 = 2A[sin(2*pi(t(f1 + f2)/2 - 12.75/334))cos(2pi(t(f1 - f2)/2 - 0.25/334))]

so the two frequencies heard are (f1 + f2)/2 and (f1 - f2)/2

f1 = 440 Hz

hence (440 - f2)/2 = 14

f2 = 440 - 28 = 412 Hz

d. length of tube, l = 0.7 m

maximum wavelength = lambda

lambda = 2*l

minimum freq = f = c/lambda = c/2l = 343/2*0.7 = 245 Hz

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