How Can You Analyze the Expression of a Traveling Wave?

[~angel~]

Consider a traveling wave described by the formula

y_1(x,t) = Asin(kx - wt).

This function might represent the lateral displacement of a string, a local electric field, the position of the surface of a body of water, or any of a number of other physical manifestations of waves.

The expression for a wave of the same amplitude that is traveling in the opposite direction is Asin(kx + wt).

The sum of these 2 waves can be written in the form y_s(x,t) = y_e(x)*y_t(t). Where y_e only depends on displacement and y_t depends on the time.

Find y_e(x) and y_t(t). Keep in mind that y_t(t) should be a trigonometric function of unit amplitude. Express your answers in terms of A, k, x, w, and t.

I know I'm meant to use the identity sin(A-B) = sinAcosB - cosAsinB, but I don't know how to apply it.

Any help would be great.

Thank you.

 

[dextercioby]

A\sin\left(kx+\omega t\right)+A\sin\left(kx-\omega t\right)

Factor "A" and use the identity

\sin\left(\alpha+\beta\right)=\sin\alpha\cos\beta+\sin\beta\cos\alpha

for

\left\{\begin{array}{cc}\alpha = kx & \beta=\omega t\\ \alpha=kx & \beta=-\omega t\end{array}\right

Daniel.

 

[~angel~]

Ok, I got A[2sinkxcoswt + 2coskxsinwt]. How do you then go from this to finding y_e(x) and y_t(t)?

 

[dextercioby]

Nope.Use the fact that

\sin \left(-\beta\right)=-\sin\beta

Post the result.

Daniel.

 

[~angel~]

Nope.Use the fact that

\sin \left(-\beta\right)=-\sin\beta

Post the result.

Daniel.

I end up with that answer and I am using that fact.

For the first bit, Asin(kx-wt) = A[sinkxcoswt+sinwtcoskx]
For the second bit Asin(kx-wt) = A[sinkxcos(-wt)-sin(-wt)coskx]

cos(-wt) = coswt
sin(-wt) = -sin(wt)

Therefore, for the second bit, A[sinkxcoswt-(-sinwt)coskx] = A[sinkxcoswt+sinwtcoskx].

I'm not sure what I'm getting wrong.

 

[dextercioby]

Nope.

A\sin\left(kx-\omega t\right)=A\sin\left(kx+\left(-\omega t\right)\right)=A\left(\sin kx\cos\omega t-\sin \omega t\cos kx\right)

Okay?

Daniel.

 

[~angel~]

So it's 2Asinkxcoswt? How do you use that to find y_e(x) and y_t(t)?

 

[dextercioby]

If add them,u'll get

y_{e}(x)y_{t}(t)=2A\sin kx\cos\omega t

Do you see which is which...??

Daniel.

 

[~angel~]

Yep, I've got it.

Thanks