P17_054.PDF

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 2 π 1

⇒

y m =3 . 2cm .

(d) Since k =2 π/λ ,wehave k =10 π rad/m. There must be a sign diﬀerence between the t and x

terms in the argument in order for the wave to travel to the right. The ﬁgure shows that at x =0,

the transverse velocity function is 0 . 050 sin 2 t . Therefore, the function u ( x, t )is

u =0 . 050 sin π

2 t

−

10 πx

with lengths in meters and time in seconds. Integrating this with respect to time yields

y =

−

2(0 . 050)

cos π

2 t

−

10 πx + C

where C is an integration constant (which we will assume to be zero). The sketch of this function

at t =2 . 0sfor0

≤

0 . 20 m is shown.

0.03

0.02

0.01

0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

–0.01

–0.02

–0.03