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Unformatted text preview: PHYSICS 218 SOLUTION TO HW 3 Created: September 12, 2004 10:42pm Last updated: September 17, 2004 10:53am 1. We are asked to express several quantities that characterize a general wave A sin( kx ωt ). (a) We know ρ s = ρ ∂ξ ∂x 1 . Inserting this into p s ( x, t ) /p = γρ s ( x, t ) /ρ leads to the final equation p s ( x, t ) = γp ∂ξ ∂x = BkA cos( kx ωt ). (1) Thus we identify P m = BkA as the coefficient of the cosine or by setting cos equal to 1, which yields the maximum value. (b) Using (a) we have ρ s ( x, t ) = ρ ∂ξ ∂x = ρ kA cos( kx ωt ) (2) and thus R m = ρ kA . (c) We get d K d V = 1 2 ρ ∂ξ ∂t 2 = 1 2 ρ A 2 ( ω ) 2 cos 2 ( kx ωt ) (3) and thus the maximum kinetic energy density is equal to 1 2 ρ ω 2 A 2 . (d) Using d U d V = 1 2 γp ∂ξ ∂x 2 = 1 2 γp A 2 k 2 cos 2 ( kx ωt ) (4) and γp = Bρ we find with k 2 = ω 2 /c 2 d U d V max = 1 2 ρ ω 2 A 2 . (5) (e) Plugging our wave into the known formula I = γp ∂ξ ∂x ∂ξ ∂t = γp ω 2 c A 2 cos ( kx ωt ) (6) When we use ωk = ω 2 /c and set the cosine equal to 1 to obtain the maximum we find I max = γp ω 2 c A 2 . (7) 2. (a) As we know from the lecture the speed of sound is given by p B/ρ = p γp /ρ . This yields c ST P = 331 m s (8) (b) 1 There was a typo on the pink sheet that was handed out in the lecture on friday. The derivative must be taken with respect to x . 1 PHYSICS 218 SOLUTION TO HW 3 (c) We first derive the equations using the formulae from problem one and plug in numbers afterwards. Doing this we have to take into account that the maximum intensity we calculated above is not the average intensity. The average intensity is the average over one period. Averaging cos 2 gives a factor of 0 . 5....
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This note was uploaded on 09/28/2008 for the course PHYS 218 taught by Professor Pollack during the Fall '04 term at Cornell.
 Fall '04
 POLLACK
 Physics, Magnetism

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