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Physics 19 Fall, 2004 Solution, Hour Exam The graders for the problems were: 1 Lowe, 2,3 Hickman For questions about the grading, see the grader by Nov. 12. Problem 1. Two identical loudspeakers are connected to the same ampli±er so that they are in phase. The speed of sound in air is 345 m/s. 3.0 m 4.0 m 5.0 m 4.0 m (a) (b) d P (a) What are the two lowest frequencies f 1 and f 2 at which the speakers in panel (a) can be driven to produce max- imum intensity at the point P ? Sound from one speaker travels 5 − 4 = 1 m further; that distance must be λ for the lowest f and 2 λ for the next lowest f .U s ing λf = v , f 1 = v/λ 1 = 345 m / s 1m = 345 Hz f 2 = v/λ 2 = 345 m / s 0 . 5m = 690 Hz (b) The wires to one speaker are now reversed so that the loudspeakers are exactly out of phase, and the distance between the speakers is increased to d as shown in panel (b). (The two speakers and the point P again form a right triangle.) Find the smallest value of d> 3
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This note was uploaded on 08/06/2008 for the course PHYS 21 taught by Professor Hickman during the Fall '07 term at Lehigh University .
- Fall '07