P17_019.PDF

19. (a) We read the amplitude from the graph. It is about 5 . 0cm.

(b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm and again

with the same slope at about x =55cm,so λ =55cm

−

3 . 6N

v =

10 − 3 kg / m

=12m / s .

(d) The frequency is f = v/λ =(12m / s) / (0 . 40 m) = 30 Hz and the period is T =1 /f =1 / (30 Hz) =

0 . 033 s.

(e) The maximum string speed is u m = ωy m =2 πfy m =2 π (30 Hz)(5 . 0 cm) = 940 cm / s=9 . 4m / s.

(f) The string displacement is assumed to have the form y ( x, t )= y m sin( kx + ωt + φ ). A plus sign

appears in the argument of the trigonometric function because the wave is moving in the negative

x direction. The amplitude is y m =5 . 0

10 − 2 m. The formula for the displacement

gives y (0 , 0) = y m sin φ . We wish to select φ so that 5 . 0

10 − 2 . The solution is

either 0 . 93 rad or 2 . 21 rad. In the ﬁrst case the function has a positive slope at x = 0 and matches

the graph. In the second case it has negative slope and does not match the graph. We select

φ =0 . 93 rad. The expression for the displacement is

10 − 2 sin φ =4 . 0

y ( x, t )=(5 . 0

10 − 2 m) sin (16 m − 1 ) x + (190 s − 1 ) t +0 . 93 .

15 cm = 40 cm = 0 . 40 m.

of the string. Thus,

10 − 2 m,the angular frequency is ω =2 πf =2 π (30 Hz) =

190 rad / s,and the angular wave number is k =2 π/λ =2 π/ (0 . 40 m) = 16 m − 1 . According to the

graph,the displacement at x =0and t =0is4 . 0