- #1

- 53

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Should I go about this by assuming...

sin(120pi*t) = sin(120pi(t + x) - 20)

Any hints appreciated

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- Thread starter redshift
- Start date

- #1

- 53

- 0

Should I go about this by assuming...

sin(120pi*t) = sin(120pi(t + x) - 20)

Any hints appreciated

- #2

Galileo

Science Advisor

Homework Helper

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You have two sines which have are functions of position and time. (actually, considering them function of time alone is sufficient)

They both have the form:

[tex]A\sin(kx-\omega t + \phi)[/tex]

Assume the first wave reaches its maximum A at time t=0 and position x=0.

Then you have to find t when the second wave reaches its max A:

[tex]A\sin(-\omega t + \phi)=A[/tex]

Where [itex]\phi[/itex] is 20 degrees expressed in radians.

They both have the form:

[tex]A\sin(kx-\omega t + \phi)[/tex]

Assume the first wave reaches its maximum A at time t=0 and position x=0.

Then you have to find t when the second wave reaches its max A:

[tex]A\sin(-\omega t + \phi)=A[/tex]

Where [itex]\phi[/itex] is 20 degrees expressed in radians.

Last edited:

- #3

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sin (-120pi*t + pi/9) = 1

so, -120pi*t + pi/9 = pi/2

and finally t = 7/2160 seconds

Many thanks

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