# Transverse sinusoidal wave on a string

1. May 24, 2008

### K3nt70

[SOLVED] transverse sinusoidal wave on a string

1. The problem statement, all variables and given/known data
The wavefunction of a transverse sinusoidal wave on a string has the form y(x,t) = A*cos(kx +omega*t + phi), where x and y are in m, t is in s and phi is the phase constant in radians. The wave has a period T = 24.2 ms and travels in the negative x direction with a speed of 28.3 m/s. At t = 0, a particle on the string at x = 0 has a displacement of 1.80 cm and travels downward with a speed of 2.05 m/s. What is the amplitude of the wave?

2. Relevant equations

y(x,t) = Acos(kx + wt + phi)

y(x,t) = A/(((x - vt)^2) + 1)

3. The attempt at a solution

Despite my attempts using these equations, i have made ZERO headway. I think im having a problem understanding where all the information in the question fits. I did a quick diagram to show what i THINK is happening (is probably not to scale). I guess i have to be PMed for anyone to see it?

2. May 24, 2008

### lzkelley

What is y in these equations? What is A?

3. May 24, 2008

### K3nt70

A is amplitude and according to my course notes, "At time t , displacement at given location x is y(x,t)."

4. May 25, 2008

### alphysicist

Hi K3nt70,

You can already find k and w, because they give you the period and wave speed. What do you get for those values?

You also need the formula for phi as a funtion of the initial displacement and velocity in the y-direction (at x=0,t=0).

Once you have k, w, and phi, you can solve for A by using your first equation.

5. May 25, 2008

### K3nt70

i solved for w fine, (259.64) but i dont know how to solve for k. I know that k = 2pi/lamda but i have no idea what lamda is in this situation..

edit: v = lamda/T gave me lamda,(0.68486) and then i solved for k using k = 2pi/lamda (9.17441). also, in order to use y(x,t) = Acos(kx + wt + phi) dont i need a value for t, time?

Last edited: May 25, 2008
6. May 25, 2008

### alphysicist

The problem gives you a value for the displacement at t=0 and x=0, so all you need now is to find phi and you can solve for A.

To find phi, your textbook should have a formula for phi in terms of the tangent of phi and the initial velocity and displacement at x=0. You have to be careful to use the right signs for the values; what do you get for that?

7. May 25, 2008

### K3nt70

i cant find any formula similar to what you described above. all the formulas i found with phi also have amplitude in them.

8. May 25, 2008

### alphysicist

You have:

$$y = A \cos (k x + \omega t + \phi)$$

If you take the derivative, you find the speed of the particles in the y direction:

$$v_y = - A \omega \sin(k x + \omega t + \phi)$$

If you set x=0, and t=0, then you can plug in the displacement and initial speed they give you. What do you get for that?

Then you can take the ratio of the two equations, and convert the sine/cosine to a tangent. You get a standard equation which is something like:

$$\frac{v_y}{y} = -\omega \tan\phi$$

where the $v_y$ and $y$ on the left hand side are the values at x=0, t=0.

9. May 25, 2008

### K3nt70

ok, so ive worked out phi to be -77.2 im going to put them into the velocity formula and solve for A...

edit: hmm my amplitude came out to be 4.305E-4 m which is wrong.

Last edited: May 25, 2008
10. May 25, 2008

### alphysicist

I don't see how you got phi to be -77.2; what numbers are you using to calculate it?

11. May 25, 2008

### K3nt70

$$\frac{2.05 m/s}{0.0018 m} = -259.64 *tan\phi$$

and solved for phi..

12. May 25, 2008

### alphysicist

There are three things: The initial velocity at x=0 is negative; your denominator is not correct (there is one too many zeros); also, it's probably best to do everything in radians (it's easiest to plug into the formulas that way).

13. May 25, 2008

### K3nt70

okay, so my new phase constant is 0.413368

so... -28.3 m/s = -A(259.64)Sin(0.413368)

i got -0.27 m for A which is wrong. Also, my calculator is now in radians :s

Last edited: May 25, 2008
14. May 25, 2008

### alphysicist

Remember that we have two speeds here; the speed on the left hand side of that equation is not the wave speed in the x-direction; it is the speed of the string in the y-direction, and so the number of the left needs to be -2.05 m/s

15. May 25, 2008

### K3nt70

YES! thanks for you're help! This led to an addition two correct questions on my assignment.