529_Physics ProblemsTechnical Physics

# 529_Physics ProblemsTechnical Physics - 531 Chapter 18...

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Chapter 18 531 P18.16 yA k x t = 2 0 sin cos ω =− 2 2 0 2 2 y x Ak kx t sin cos 2 2 0 2 2 y t x t ωω sin cos Substitution into the wave equation gives −= F H G I K J 2 1 2 0 2 2 0 2 t v x t sin cos sin cos ej This is satisfied, provided that v k = P18.17 yx t 1 300 0600 =+ .s i n . π af cm; t 2 i n . cm yy y x t x t t =+= + = 12 3 00 0 600 3 00 0 600 600 i n c o s . i n c o s . i n c o s . ππππ ππ bg b g g g cm cm (a) We can take cos . 1 t bg = to get the maximum y . At x = 0250 . c m , y max i n . . == 6 00 0 250 4 24 cm cm a f a f (b) At x = 0500 c m , y max i n . . 6 00 0 500 6 00 cm cm a f (c) Now take cos . 1 t to get y max : At x = 150 c m , y max i n . . = 6 00 1 50 1 6 00 cm cm a f a fa f (d) The antinodes occur when x n = λ 4 n = 135 ,,, But k 2 , so = 200 c m and x 1 4 . cm as in (b) x 2 3 4 . cm as in (c) x 3 5 4 250 c m P18.18 (a) The resultant wave is k x t F H G I K J F H G I K J 2 22 sin cos φ The nodes are located at kx n += 2 so x n kk πφ 2 which means that each node is shifted
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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