problem15_54

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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15.54: a), b) c) The displacement is a maximum when the term in parentheses in the denominator is zero; the denominator is the sum of two squares and is minimized when , vt x = and the maximum displacement is A . At cm, 4.50 = x the displacement is a maximum at s. 10 2.25 ) s m (20.0 m) 10 (4.50 3 2 - - × = × = t The displacement will be half of the maximum when s. 10 2.75 and s 10 1.75 ) ( or , ) ( 3 3 2 2 - - × × = ± = = - v A x t A vt x d) Of the many ways to obtain the result, the method presented saves some algebra and minor calculus, relying on the chain rule for partial derivatives. Specifically, let , ) , ( vt x t x u u - = = so that if . and ), ) , ( v du dg t u du dg t f du dg
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Unformatted text preview: t u du dg x f g(u t x f-= ∂ ∂ = ∂ ∂ = ∂ ∂ = ∂ ∂ = (In this form it may be seen that any function of this form satisfies the wave equation; see Problem 15.59.) In this case, , ( ) , ( 1 2 2 3-29 + = u A A t x y and so 3 2 2 2 2 3 2 2 2 2 2 3 ) ( ) 3 ( 2 , ) ( 2 u A u A A x y u A u A x y +--= ∂ ∂ +-= ∂ ∂ , ) ( ) 3 ( 2 , ) ( 2 2 2 2 2 2 3 2 2 2 2 2 2 3 u A u A A v t y u A u A v t y +--= ∂ ∂ + = ∂ ∂ and so the given form for ) , ( t x y is a solution to the wave equation with speed . v...
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