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The textbook gives it in the second form. I think that using one over the other gives the same answer but in the opposite sign. I would like to know which one to use over the other. Any clarification is appreciated.

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- Thread starter DannyPhysika
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- #1

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The textbook gives it in the second form. I think that using one over the other gives the same answer but in the opposite sign. I would like to know which one to use over the other. Any clarification is appreciated.

- #2

jtbell

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- #4

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Tbh it's also frustrating that I find 3-4 clear mistakes on every page of the teacher's notes. I much prefer when the teacher lets the students know the exact sections from the official textbook that are covered because at least I trust the textbook.

- #5

nasu

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[tex]Asin(\omega t - kx +\pi)[/tex]

so it's a phase difference of 180 degrees.

In general you can have any initial phase:

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

The value of the phase can be found from the initial configuration of the wave. Or we can take as "initial" the configuration that gives phase of zero (or 180 degree).

Could you show a problem for which your answer is not as in the book?

- #6

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a) A certain string has a linear mass density of 0.25 kg/m and is stretched with a tension of 25 N. One end is given a sinusoidal motion with frequency 5 Hz and amplitude 0.01m. At time t=0, the end has zero displacement and is moving in the positive y-direction. Find the wave speed, amplitude, angular frequency, period, wavelength, and wave number.

I found all of those, and then I am asked for:

b) Write the wave function describing the wave.

And my answer, based on the textbook, is: Asin(kx - ωt) (I'm using the symbols rather than the actual values to simplify things.

But the answer key is: Asin(ωt - kx)

This leads to opposite answers in the subsequent parts c and d of the question. For example:

d) Find the transverse velocity of the point, x=0.25 m at time t=0.1

so I write my answer as v = -Aωcos(kx - ωt) because I take the derivative of the textbook function with respect to t, and I get the answer 0.22 m/s, but the answer key gives it as -0.22 m/s.

- #7

jtbell

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At time t=0, the end has zero displacement and is moving in the positive y-direction.

[snip]

And my answer, based on the textbook, is: Asin(kx - ωt)

At x = 0 your answer reduces to A sin (-ωt). At t = 0 this obviously gives a displacement of zero. A short time later, does it give a positive displacement as the problem statement implies?

- #8

nasu

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The velocity of a point is given by the derivative of the displacement in respect to time.

If we start with the format

[tex]y(x,t)=Asin(\omega t -kx +\phi)[/tex]

then

[tex]v(x,t)=\omega A cos(\omega t -kx +\phi)[/tex]

At t=0, x=0,

[tex]y=A sin(\phi)=0[/tex]

This can happen for a phase of 0 or 180.

For

[tex]v=\omega A cos(\phi)[/tex]

to be positive you need phase zero so the equation will be

[tex]y(x,t)=Asin(\omega t -kx)[/tex]

If we start with the format

[tex]y(x,t)=Asin( kx -\omega t+\phi)[/tex]

The velocity will be

[tex]v(x,t)=-\omega A cos(kx -\omega t +\phi)[/tex]

The condition fo v(0,0) to be positive will require a phase of 180 so the final formula will be

[tex]y(x,t)=Asin( kx -\omega t+180)[/tex] which is the same as

[tex]y(x,t)=Asin(\omega t -kx)[/tex]

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