Answer:
a) [tex] y(x,t=0)= 2cm cos(2\pi x -4\pi *0) = 2cm cos(2\pi x)[/tex]
And we can see the plot in the first figure attached.
b) [tex] y(x,t=1/8)= 2cm cos(2\pi x -4\pi *\frac{1}{8}) = 2cm cos(2\pi x-\frac{\pi}{2})[/tex]
And we can see the result on the second figure attached.
As we can see we have a translation on the x axis for this new function.
c) [tex]\frac{dx}{dt}=-8\pi cm sin(2\pi x -4\pi t)[/tex]
Explanation:
For this case we have the following function given (assumed):
[tex] y(x,t) = 2 cm cos (2\pi x -4\pi t)[/tex]
Where x is in cm and t in seconds
Part a
For this case we need to replace the value of t =0 and we got:
[tex] y(x,t=0)= 2cm cos(2\pi x -4\pi *0) = 2cm cos(2\pi x)[/tex]
And we can see the plot in the first figure attached.
Part b
For this case we just need to replace the value of t =1/8 s and we have the following function:
[tex] y(x,t=1/8)= 2cm cos(2\pi x -4\pi *\frac{1}{8}) = 2cm cos(2\pi x-\frac{\pi}{2})[/tex]
And we can see the result on the second figure attached.
As we can see we have a translation on the x axis for this new function.
Part c
The velocity on this case is given by the first derivate of the position respect to the x axis and we got:
[tex] \frac{dx}{dt}= -2cm * (4\pi) sin (2\pi x -4\pi t)[/tex]
[tex]\frac{dx}{dt}=-8\pi cm sin(2\pi x -4\pi t)[/tex]
