handout07 - ∂ 2 p ∂t 2 = c 2 ∇ 2 p find and...

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PHYSICS 218 SOLUTION TO HANDOUT 7 Created: September 25, 2004 10:44pm Last updated: September 25, 2004 11:47pm Irrotational Acoustic Waves and Velocity Potential For a plane acoustic wave, we know that the fluid displacement % and velocity ˙ % satisfy ∇ × % = 0 = ∇ × ˙ % . We also know from vector analysis that given a scalar function ϕ , we always have ∇ × ∇ ϕ = 0. It is tempting to express the fluid velocity as a gradient of a scalar ˙ % = ϕ . ϕ is known as the velocity potential. 1. Using the fact that the fluid pressure can be described by p = - B ∇ · % , where B is the bulk modulus, find ∂p ∂t and 2 p ∂t 2 in terms of ϕ . Solution: Taking partial derivative with respect to time of the equation p = - B ∇ · % , we find ∂p ∂t = - B ∂t ( ∇ · % ) = - B ∇ · ˙ % = - B 2 ϕ . Taking partial derivative with respect to time again, we have 2 p ∂t 2 = - B 2 ∂ϕ ∂t . 2. Now, exploit the fact that the pressure satisfies the wave equation
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Unformatted text preview: ∂ 2 p ∂t 2 = c 2 ∇ 2 p , find and expres-sion for p in terms of ϕ . Solution: Using the wave equation, we have c 2 ∇ 2 p =-B ∇ 2 ∂ϕ ∂t . From this, we then guess that p =-B c 2 ∂ϕ ∂t . 3. From the expression you found for p , take partial derivative with respect to time and compare the resulting expression with your previous expression for ∂p ∂t . Does ϕ satisfy the wave equation? If so, what is the wave velocity for ϕ ? Solution: Now take the partial derivative with respect to time of p =-B c 2 ∂ϕ ∂t , we find ∂p ∂t =-B c 2 ∂ 2 ϕ ∂t 2 . Comparing this with the expression we had earlier for ∂p ∂t =-B ∇ 2 ϕ , we conclude that ∂ 2 ϕ ∂t 2 = c 2 ∇ 2 ϕ . So ϕ satisfies the wave equation with the same velocity c . 1...
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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.

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