54. (a) The frequency is f =1 /T =1 / 4Hz ,so v = fλ =5 . 0cm/s . (b) We refer to the graph to see that the maximum transverse speed (which we will refer to as u m )is 5 . 0 cm/s. Recalling from Ch . 12 the simple harmonic motion relation u m = y m ω = y m 2 πf ,wehave 5 . 0= y m µ 2 π 1 4 ¶ = ⇒ y m =3 . 2cm . (c) As already noted, f =0 . 25 Hz. (d) Since k =2 π/λ ,w ehav e k =10 π rad/m. There must be a sign diFerence between the t and x terms in the argument in order for the wave to travel to the right. The ±gure shows that at x =0, the transverse velocity function is 0
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