- #1

hotmail590

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- 0

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

Thank you for your help!

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- Thread starter hotmail590
- Start date

- #1

hotmail590

- 52

- 0

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

Thank you for your help!

- #2

- 6,636

- 2,009

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

- #3

hotmail590

- 52

- 0

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

Asin(kx+wt) = AsinkxAcoswt +AcoskxAsinwt

But wouldn't that equal to zero?

- #4

- 6,636

- 2,009

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

hotmail590

- 52

- 0

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

hotmail590

- 52

- 0

anyone please?

- #7

Galileo

Science Advisor

Homework Helper

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- #8

hotmail590

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- 0

- #9

- 6,636

- 2,009

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

hotmail590

- 52

- 0

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

- #11

- 6,636

- 2,009

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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