- #1

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I have no clue how to add 2 sins with diffent phases.

Thank you for your help!

- Thread starter hotmail590
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- #1

- 52

- 0

I have no clue how to add 2 sins with diffent phases.

Thank you for your help!

- #2

- 5,751

- 1,048

Can you re-express sin(kx+wt) using a trig identity?

- #3

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so Asin(kx-wt) = -AsinkxAcoswt -AcoskxAsinwt

Asin(kx+wt) = AsinkxAcoswt +AcoskxAsinwt

But wouldnt that equal to zero?

- #4

- 5,751

- 1,048

Your identity is fine.

Check your use of it. Check the signs you wrote down.

Check your use of it. Check the signs you wrote down.

- #5

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oh i saw where i made my mistake

the answer should be 2Asin(kx)cos(wt)?

Now after I get that how would I find the

ye(x), called the envelope, depends only on position

and yt(x) depends only on time

yt(x) should be a trigonometric function of unit amplitude.

I need to express ye(x) and yt(x) in terms of A k omega x and t.

Is there partial derviative involved?

Thank you for your help!!

the answer should be 2Asin(kx)cos(wt)?

Now after I get that how would I find the

ye(x), called the envelope, depends only on position

and yt(x) depends only on time

yt(x) should be a trigonometric function of unit amplitude.

I need to express ye(x) and yt(x) in terms of A k omega x and t.

Is there partial derviative involved?

Thank you for your help!!

Last edited:

- #6

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anyone please?

- #7

Galileo

Science Advisor

Homework Helper

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- #8

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- #9

- 5,751

- 1,048

Plot your superposed function for (say) t=0.01. Notice where the x-positions of the peaks are now.

Which way did it shift? If 0.01 is too complicated, try t=(1/12)*(2*pi/w).

If it's not clear, try doubling the value of t you just used.

- #10

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I graphed sin(x)cos(0.01) and then sin(x)cos((1/12)*(2pi))

- #11

- 5,751

- 1,048

The second expression is easy to plot a graph of

"amplitude sin(x)cos((1/12)*(2pi)) [at time (1/12)*(2pi)] vs position x".

Can you numerically evaluate cos((1/12)*(2pi)), where pi=[itex]\pi[/itex].

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