A Sinusodial wave is propagating

1. Jan 20, 2012

Firben

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? (solved)
B) What is the period of the wave? (solved)
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. (solved)

If the wave is moving in the -x-direction, determine the wavelength and wave speed. What should i do here ?

http://s716.photobucket.com/albums/...rent=5508afb2-b0b2-404b-aa14-09f9c8e37e4a.jpg

2. Jan 20, 2012

Simon Bridge

Exactly the same as before except the velocity is the other way.
What difference does the direction the wave travels make?

3. Jan 20, 2012

Firben

Yes, but how can it be 6.0 m/s?
Thats is whats stands in the answer sheet

Last edited: Jan 20, 2012
4. Jan 20, 2012

Firben

5. Jan 20, 2012

Simon Bridge

this cannot possibly be correct
- when you changed units from m to mm you forgot to multiply by 1000
- you cannot get length/time from multiplying a length and a time.
... try again.

show me how you worked out A, B, and C - what is your reasoning?

6. Jan 20, 2012

Firben

A) A=4mm
B) T=0.040s (red wave)
C) it should be 3.6 m/s
What should i do to find v?

Last edited: Jan 20, 2012
7. Jan 20, 2012

vela

Staff Emeritus
Pick a peak on the x=0 wave and, from the diagram, figure out how long it takes to reach x=0.090 m. The speed of the wave will then be the distance that point traveled, 0.090 m, divided by the time it took to get there.

8. Jan 20, 2012

Simon Bridge

Wot he said - easier to find a place where y=0, but you also need the same slope.
You just need the position of the wave at two different times.
speed is distance over time :)

To be clear - and because this is not always taught:

An if arbitrary distortion of the string is: y=f(x)
If it propagates to the right (+x) direction at speed v, then the equation becomes y(x,t)=f(x-vt)

In this case f(x)=Asin(kx): k=2π/λ ...(k has the effect of turning the distance x into an angle)

Therefore: y(x,t)=Asin[k(x-vt)]=Asin(kx-ωt). You can rewrite this in terms of λ and v - it will be more useful to this problem that way.

In your case this translates into the time-waves at two different x's: y(0,t) and y(0.0900,t) ... which are the red and blue lines respectively.

generally: y(x,t)= 4.sin(kx-2πt/0.04)

so: y(0,t)=-4sin(2πt/0.04) and y(0.09,t)=4.sin(0.09k - 2πt/0.04)

it is a good idea to keep the pi's separate for as long as you can.
you can use the graph to solve for k in the second one and so get the wavelength.
and get the speed from v=fλ ... but it is easier to do distance over time from the graph.

changing the direction of travel is a matter of changing the sign of v in the derivation.

Last edited: Jan 20, 2012
9. Jan 24, 2012

Firben

0.09k=2pi*t/0.04
What is the value of t?

10. Jan 24, 2012

Simon Bridge

woah: where did that come from?
Is this still the same problem?
Did you try out post #7 first?

11. Jan 26, 2012

Firben

Yes. How can i find the spped from the graph ?

12. Jan 27, 2012

Simon Bridge

The two graphs show you the height-vs-time of different positions on the string.
These graphs are due to a wave f(x) moving in the x direction.
So we needs some way to find out how long it takes f(x) to move a fixed distance.
The distance we have is 0.09m - all the points in f(x) move at the same speed, so we just need to know the time one point in f(x) crosses x=0, and the time the same point in f(x) crosses x=0.09. An easy to identify point is when f(x) is a maximum.

So look at the time when the x=x0=0 graph first reaches maximum - record that at t0
Then look for the time the x=x1=0.09m graph first reaches it's maximum - record it as t1
t1-t0 is how long it took for f(x) to move from x0 to x1.
Now you have a distance, and the time it took to travel that distance - you can find the speed.

(Note: you don't have to use peaks - I used the points when the graph crosses from -y to +y - easier to pinpoint but also easier to stuff up.)

The method in post #8 involves two simultaneous equations and two unknowns.