Question

4. In a large lecture hall, the professor uses a microphone and her voice is amplified over two speakers located at the front of the fall in the upper right and left corners. The speakers are separated by a distance of D-15 meters. You pick a spot in the hall that is located Si-13.0 meters from the right speaker and S 12.5 meters from the left speaker. Assume the speed of sound is 343 m/s. Assuming the speakers are connected with the same phase which frequencies will be canceled out, or at least minimized? Which frequencies will be maximized? As part of the lecture demonstration, the professor connects separate frequencies to each of the speakers. The right speaker puts out at tone of 440 Hz. The left speaker puts out a lower frequency, and you hear a beat frequency of 14 Hz. What is the frequency put out by the left speaker? After the lecture is over, the professor forgets to turn off the speakers and they are both sending out the same pure tone with frequency 6860 Hz. As you leave the lecture hall, you pass between the two speakers, moving directly towards one speaker and away from the other one. You hear a beat frequency of 120 Hz. How fast are you walking? a. b. c. d.

Answer 1

(a) For minimum Delta x = (2 n - 1) lambda/2 n = 1, 2, 3,... s_1 - s_2 = (2 n - 1) v/2 f Rightarrow f = (2 n - 1) v/2 (s_1 - s_2) f = (2 n - 1) 343/2 times 0.5 Hz f = (2 n - 1) 343 Hz Thus frequencies which will cancel out are f = 343 Hz, 1029 Hz, 1715 Hz, ... (b) for frequencies to be maximized f = n v/(s_1 - s_2) = n times 343/0.5 Hz n = 1, 2, 3, ... f = 686 Hz, 1372 Hz, 2058 Hz, ... (c) let frequency of left speaker be f_L be f_L and frequency of right speaker be f_R Therefore Beat frequency, f_beat = f_R - f_L 14 = 440 - f_L Rightarrow f_L = (440 - 14) Hz f_L = 426 Hz leftarrow frequency of left speaker (d) ket V_0 be the speed of walking then, f_beat = (v + v_0/v - v - v_0/v) f_0 = 2 v_0/v f Therefore 120 = 2 v_0/343 times 6860 Rightarrow v_0 = 3 m/s