Two loudspeakers S_1 and S_2, 2 m apart, emit the same single-frequency tone in

Question

Two loudspeakers S_1 and S_2, 2 m apart, emit the same single-frequency tone in phase at the speakers. A listener L is located directly in front of speaker s_1 so that the lines LS_1 and S_1 S_2 are perpendicular. L notices that the intensity is at a minimum when L is 6 m from speaker S_1. What is the lowest possible frequency of the emitted tone? The speed of sound in air is 340 m/s. A) 1270 Hz B) 254 Hz C) 381 Hz D) 1016 Hz E) 508 Hz A wave on a string has a frequency of 100Hz and travel at a speed of 23 m/s. The minimum distance between two points with a phase difference of 60 degree is A) 0.31 m B) 0.04 m C) 0.08 m D) 0.15 m E) 0.02 m A string 4m long has a mass of 0.046kg. When fixed at both ends, it vibrates with a fundamental frequency of 160 Hz. The speed of a transverse wave in the string is A) 0.025 m/s B) 0.32km/s C) 40.0 m/s D) 0.64 hm/s E) 1.28 km/s

Explanation / Answer

2)

wavelength lambda = v/f = 23/100 = 0.23 m

path difference = (lambda/(2pi))*phase difference

path difference = (0.23/(2pi))*(pi/3) = 0.04 m <<<--answer

3)

fundamental frequency f = v/2L

160 = v/(2*4)

v = 1280 m/s

1 m = 10^-3 km

speed v = 1.28 km/s

option E

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