A function has the form y(x,t)= e^(-1-(t+x)^2).

a) Fixing the time at t=0, t=1, t=2, draw y as a function of x.
b) Is this a wave? Determine the wave number k and circular frequency .
c) What is the propagation velocity? Is it positive or negative?
d) In what direction is it propagating (+x or –x)?

Question

A function has the form y(x,t)= e^(-1-(t+x)^2).

a) Fixing the time at t=0, t=1, t=2, draw y as a function of x.
b) Is this a wave? Determine the wave number k and circular frequency .
c) What is the propagation velocity? Is it positive or negative?
d) In what direction is it propagating (+x or –x)?

anonymous · Answer

@saifoo.khan  @hartnn @dpaInc @JamesJ @cwrw238 @robtobey @radar

anonymous · Answer

thank you, i need help with b and c,

I drew (a), and d I think the answer is that it is not moving

anonymous · Answer

You can say this is a wave because the function has the argument form $$x \pm v$$ so x and t are combined in a linear fashion. You could also actually see if this is a solution of the partial differential equation of a wave but either argument is valid.

so the argument of the function is called the phase which in this case is $$\phi =-1-(t+x)^2)$$ so you are looking for the change of position with respect to time so you get
 $$0 =-(2tdt+(xdt+tdx)+2xdx)$$ since the you are assuming the phase is constant or x and t are such that the phase is constant so get dt on one side and dx on the other

$$2tdt+xdt = -tdx-2xdx$$ factor out dx and dt
$$dt(2t+x) = (-t-2x)dx$$ finally isolate dx/dt
$$dx/dt =-(2t+x)/(t+2x)$$ would be a function for the velocity. So in this case it is moving in the negative x-direction as that whole function is negative.

anonymous · Answer

yes but still I do not get what would be the wave number k and w the angular frequency in this case? would they be as follow, k=1rad/m, w=1rad/s

anonymous · Answer

@saifoo.khan  @hartnn @dpaInc @JamesJ @cwrw238 @robtobey @radar