Assignment 2 Solution

Assignment 2 Solution - 5 Hz. The intensity I = 1 2 p 2 c ....

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PHYS-232 Heat and Waves - Homework 2 Winter Semester 2008 Due on Thursday, January 31, 2008 1. y ( x,t ) = Asin ( kx - wt ). kx AB = π/ 6, where x AB = 0 . 02 m . We can get k = π/ 6 / 0 . 02 = 100 π/ 12. The wave length of the plane sinusoidal wave is: λ = 2 π/k = 0 . 24 m . The wave speed is: c = = 10 × 0 . 24 = 2 . 4 m/s 2. We first calculate 2 z ∂t 2 , using z = Acos ( kx - wt ). 2 z ∂t 2 = - Aw 2 cos ( kx - wt ) (1) We then calculate 2 z ∂x 2 . 2 z ∂x 2 = - Ak 2 cos ( kx - wt ) (2) If c = q Y/ρ , then w/k = c = q Y/ρ and w 2 = k 2 * Y/ρ . Obviously, under this condition, 2 z ∂t 2 = Y ρ 2 z ∂x 2 (3) 3. From p ( x,t ) = 5 × 10 - 3 Pa · sin ( x 3 m - wt ), you can obtain k = 1 3 m . The wave length λ = 2 π k = 6 π m. We know the speed of sound is c = 330 m/s . Frequency f = c/λ = 17 .
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Unformatted text preview: 5 Hz. The intensity I = 1 2 p 2 c . p = 5 10-3 Pa and = 1 . 29 kg/m 3 , so I = 2 . 9 10-8 w/m 2 . Intensity level= 10 log I I = 44 . 7 db . 4. The velocity c = q T/ . The wave length can be 2 L/n , where n = 1 , 2 , 3 ... . The frequency f = c/ = n 2 L q T . 5. I = 1 2 p 2 c . The ratio R = I 1 I 2 = p 2 01 p 2 02 . From p = Pkz and k = 2 / = 2 f/c , we calculate R = k 2 1 z 2 01 k 2 2 z 2 02 = ( f 1 z 01 ) 2 ( f 2 z 02 ) 2 = 4 9 1...
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This note was uploaded on 04/29/2008 for the course PHYS 232 taught by Professor Vinals during the Winter '08 term at McGill.

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