P17_050.PDF

50. From the x = 0 plot (and the requirement of an antinode at x = 0), we infer a standing wave function

of the form

(0 . 04)cos( kx )sin( ωt )whe ω = 2 π

y =

−

= π rad / s

with length in meters and time in seconds. The parameter k is determined by the existence of the node

at x =0 . 10 (presumably the ﬁrst node that one encounters as one moves from the origin in the positive

x direction). This implies k (0 . 10) = π/ 2sothat k =5 π rad/m.

(a) With the parameters determined as discussed above and t =0 . 50 s, we ﬁnd

y =

0 . 04 cos( kx )sin( ωt )=0 . 04 m at x =0 . 20 m .

(b) The above equation yields zero at x =0 . 30 m.

u = dy

dt =

0 . 04 ω cos( kx )cos( ωt )=0 at t =0 . 50 s

0 . 126m / sat t =1 . 0s.

(e) The sketch of this function at t =0 . 50 s for 0

−

≤

0 . 40 m is shown.

0.04

0.02

0.1

0.2

0.3

0.4

–0.02

–0.04

−

where x =0 . 20 m.

(d) The above equation yields u =