Y displacements in sine wave at (x) and (x + 2 wavelengths)

[Point Conception]

Homework Statement

A sine wave in + x direction with max amplitude of 1
wavelength= 2.85 m
wavenumber, k = 6.28/2.85 m = 2.2 rad/m
w=8rad/m
frequency=1.27 cyc/sec
vel.= 3.63 m/sec
at x = 20m t= 5.5 sec
at x + 2 wavelengths x= 25.7m t=7.07 sec
by definition the y displacement is the same at (x) as it is at (x + 2 wavelengths)

Homework Equations

y (x,t)= A sin(kx-wt)

The Attempt at a Solution

just considering y= Asin(kx) at x= 20m ,y= .0177
with y= Asin(k x+2 wavelengths) x= 25.7m, y=-.00866
with radian set calculator
I don't understand wht the y displacements are not the same

Also if the other variables are put into the solution A=Aosin(kx-wt)
then in both cases (kx-wt) =0 ?

Homework Equations

The Attempt at a Solution

 

[rl.bhat]

equation of a moving in +x direction can be written as
y(x,t)= yo*sin2π(t/T - x/λ)
Here x = 20 m. It can be written as n*λ + x'. Similarly t can be written as n*T + t'
So 20 = 7*2.85 + x'. Or x' = 0.05 m.
T = 2π/8 = 0.785 s.
t' = T/λ*x' =...?
Rewrite the equation as
y(x,t)= yo*sin2π(t'/T - x'/λ) and solve for y.

 

[Point Conception]

thanks i cleared up my confusion with the correct values; with y=yo(sin2pi) x/wl -t/T
at t=0
x=20m
wavelengh =2.85m
T=.785s
y=yo(sin2pi) x/wavelength y= sin44.07= .087
then after wave travels two wavelengths
x=25.7 m
t=2T = 1.57s

y=yo (sin2pi) 25.7m/2.85m - 1.57s/.785s =sin44.07=.087 checks out with above

actually in your reply above i don't see how x' = .05m since two wavelenths is 5.7m

 

[rl.bhat]

As I have indicated, the equation of wave moving in the positive x direction is
y = yo*sin[2π(t/T - x/λ)]
I presume, in the problem x , velocity. frequency, wavelength and yo is given.
So the dispalcement y at x = 20 m can be found by
y = sin[2π(5.5/0.785 - 20/2.85)]
Now find y.