P17_007 - 0 cm sin 62 8 m − 1 x − 2510 s − 1 t(b...

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7. (a) We write the expression for the displacement in the form y ( x, t )= y m sin( kx ωt ). A negative sign is used before the ωt term in the argument of the sine function because the wave is traveling in the positive x direction. The angular wave number k is k =2 π/λ =2 π/ (0 . 10 m) = 62 . 8m 1 and the angular frequency is ω =2 πf =2 π (400 Hz) = 2510 rad / s. Here λ is the wavelength and f is the frequency. The amplitude is y m =2 . 0cm. Thus y ( x, t )=(2
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Unformatted text preview: . 0 cm) sin (( 62 . 8 m − 1 ) x − ( 2510 s − 1 ) t ) . (b) The (transverse) speed of a point on the cord is given by taking the derivative of y : u ( x, t ) = ∂y ∂t = − ωy m cos( kx − ωt ) which leads to a maximum speed of u m = ωy m = (2510 rad / s)(0 . 020 m) = 50 m / s. (c) The speed of the wave is v = λ T = ω k = 2510 rad / s 62 . 8 m − 1 = 40 m / s ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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