# Finding wavelength and wave speed in a sinusoidal wave.

## Homework Statement

A sinusoidal wave is propagating along a stretched string that lies along the x-axis. The displacement of the string as a function of time is graphed in the figure (attachment) for particles at x=0m and x=0.0900m.
(A) What is the amplitude of the wave?
4mm
(B) What is the period of the wave?
0.04s

I am confused by the following questions:
(C) You are told that the two points x=0 an x=0.09m are within one wavelength of each other. If the wave is moving in the +x-direction, determine the wavelength and wave speed.
ANSWER KEY: 0.14m and 3.5m/s
(D) If the wave is moving in the -x-direction, determine the wavelength and wave speed.
ANSWER KEY: 0.24m and 6.0m/s

## Homework Equations

y(x,t)=Acos(kx +/- ωt)
v=λf

## The Attempt at a Solution

Initially I thought the wavelength would be 0.09 since it says 0 and 0.09 are one wavelength away from each other... But then I looked at the answer key and it seems my interpretation is wrong. Does anyone know what the question is asking in C?

## The Attempt at a Solution

#### Attachments

• 5508afb2-b0b2-404b-aa14-09f9c8e37e4a.jpeg
12 KB · Views: 1,696

## Answers and Replies

Delphi51
Homework Helper
It doesn't say 0 and .09 are one wavelength apart.
If they were, they would peak simultaneously.
It looks like one peaks about .025 s after the other. That would mean the wave went one wavelength in time .025. Can you get a wavelength or velocity out of that? If you get one, you can find the other with the wave equation.

Spinnor
Gold Member
It doesn't say 0 and .09 are one wavelength apart.
If they were, they would peak simultaneously.
It looks like one peaks about .025 s after the other. That would mean the wave went one wavelength in time .025. Can you get a wavelength or velocity out of that? If you get one, you can find the other with the wave equation.

I don't think that is correct. Depending on which way the wave is going the velocity will be about

.09m/.025s or .09m/.015s

?

Confusing graph at first.

I don't think that is correct. Depending on which way the wave is going the velocity will be about

.09m/.025s or .09m/.015s

?

Confusing graph at first.

Could you please explain to me how you got 0.15s for the -x direction? Isn't the time interval the same?

But yeah I think he is correct, it took 0.025s to travel from 0m to 0.09m, so the velocity is about 3.6m/s and wavelength is 0.144m.

Oh I see because we start at x=0.

Thanks for the help!

Im a little confused on why the wavelength is different when the wave is moving in the -x direction. Can somebody shed a little light in this for me?

Spinnor
Gold Member
Im a little confused on why the wavelength is different when the wave is moving in the -x direction. Can somebody shed a little light in this for me?

It is a confusing ( to me) graph. You have two seperate observation points with the data graphed. There is ambiguity from the graph as to which way the wave is traveling, you can't tell from the graph. But once they tell you which way the wave is going you can determine the wavelength and the velocity. Let x =0 = a and x = .09 = b. If the wave moves to the right the wave peaks at a then latter at b. if the wave moves to the left it peaks at b and then latter at a. From the graph you can then determine the time it takes for the wave to move .09 meters. From the graph you also know the frequency of the wave. You have enough information to figure the wave velocity.

Hope this was helpful. Good luck.

Spinnor
Gold Member
Im a little confused on why the wavelength is different when the wave is moving in the -x direction. Can somebody shed a little light in this for me?

The graphs might be of waves in some moving medium, like waves in a flowing river. Water waves move faster down stream then up-stream when viewed from the river bank.