P18_023 - n ranges from 1 to 28 and f = 1029, 1715, . . . ,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
13. Let L 1 be the distance from the closer speaker to the listener. The distance from the other speaker to the listener is L 2 = p L 2 1 + d 2 ,whe re d is the distance between the speakers. The phase diFerence at the listener is φ =2 π ( L 2 L 1 ) ,where λ is the wavelength. (a) ±or a minimum in intensity at the listener, φ =(2 n +1) π ,wh e r e n is an integer. Thus λ = 2( L 2 L 1 ) / (2 n + 1). The frequency is f = v λ = (2 n +1) v 2 ³ p L 2 1 + d 2 L 1 ´ = (2 n + 1)(343 m / s) 2 ³ p (3 . 75 m) 2 +(2 . 00 m) 2 3 . 75 m ´ =(2 n + 1)(343 Hz) . Now 20 , 000 / 343 = 58 .
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n ranges from 1 to 28 and f = 1029, 1715, . . . , 19550 Hz. (b) ±or a maximum in intensity at the listener, φ = 2 nπ , where n is any positive integer. Thus λ = (1 /n ) ³ p L 2 1 + d 2 − L 1 ´ and f = v λ = nv p L 2 1 + d 2 − L 1 = n (343 m / s) p (3 . 75 m) 2 + (2 . 00 m) 2 − 3 . 75 m = n (686 Hz) . Since 20 , 000 / 686 = 29 . 2, n must be in the range from 1 to 29 for the frequency to be audible and f = 686, 1372, . . . , 19890 Hz....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online