Question

Help needed please. The period of vibration of this wave is t. Sketch the stationary wave 0.25t after the instant shown below.

Answer 1

@experimentX

Answer 2

depends on what your period of vibration is.

Answer 3

woops!! 0.25 t given.

Answer 4

Yes.

Answer 5

Plot[{Sin[ x] + Cos[x],
Sin[ x + 0.25*2 Pi] + Cos[x + 0.25*2 Pi]}, {x, 0, 2 Pi}]

Answer 6

the first is at 0, the second is at 0.25 t ... assume the period is 1.

Answer 7

http://www.wolframalpha.com/input/?i=Plot%5B%7BSin%5B+x%5D+%2B+Cos%5Bx%5D%2C++++Sin%5B+x+%2B+0.25*2+Pi%5D+%2B+Cos%5Bx+%2B+0.25*2+Pi%5D%7D%2C+%7Bx%2C+0%2C+2+Pi%7D%5D

Answer 8

But the answer says the graph should be a straight horizontal line. O.o @experimentX

Answer 9

woops sorry!! you are right it's stationary wave.

Answer 10

Plot[{Sin[x + Pi/2] + Sin[-x + Pi/2]}, {x, 0, 2 Pi}]
after 0.25 T,
Plot[{Sin[x + Pi] + Sin[-x + Pi]}, {x, 0, 2 Pi}]

Answer 11

Can you please explain the logic? I won't have wolfram alpha in the exam. :/

Answer 12

The general equation for a stationary wave is:\[y=y _{0}\cos(2\pi ft)\sin(kx)\]where f is the frequency; t is time; and k is the wave number. Writing the equation in terms of period, T, and wavelength we get:\[y=y _{0}\cos(\frac{ 2\pi t }{ T })\sin(\frac{ 2\pi x }{ \lambda })\]Now in your problem t=0.25T so insert that into the above equation:\[y=y _{0}\cos(\frac{ 0.25*2*\pi T }{ T })\sin(\frac{ 2\pi x }{ \lambda })=y _{0}\cos(\frac{ \pi }{ 2})\sin(\frac{ 2\pi x }{ \lambda })\]Now you should see that the position, x, has no influence on y at t=0.25T.

Answer 13

Okay. Thank you. How will I determine the point for t=0.5T?
The equation would be like this:
\[y= y _{0} \cos (\pi) \sin (\frac{ 2\pi x }{ \lambda })\]
cos 180 is -1. But how do I replace values of x to draw the exact curve?