# Homework Help: String with weight attached

1. Oct 29, 2007

### azila

1. The problem statement, all variables and given/known data
A 1.50 m string of weight 1.24 N is tied to the ceiling at its upper end, and the lower end supports a weight W. When you pluck the string slightly, the waves traveling up the string obey the equation y(x,t) = (8.50 mm)cos(172m^-1x - 2730s^-1t)
How much time does it take a pulse to travel the full length of the string?
What is the weight W?
How many wavelengths are on the string at any instant of time?

2. Relevant equations
Acos2pi(x/wavelength - t/T)
Mu = Mass/L
f= 1/T
v=wavelength * f

3. The attempt at a solution
Ok, so I found the mass of the string by doing w=mg and it was .127 kg. Then I found the mu, and that was .085kg/m. The wavelength was given to us in the equation to be 172 m and the T (period) was also given in the equation to be 2730 s. So, I found the frequency which is 1/T and it was .000366. Then I found the v, speed through the equation and it was .063 m/s. So, now that I have this, I don't know how to respond to the question though, how would I find the time it takes to travel down the string? Could I just do length divided by the velocity and I don't know how to find weight of the object? If anyone can help, please help.. Thanks in advance.

2. Oct 29, 2007

### G01

You are correct in how you would find the time it takes to travel down the string.

In regards to your numbers for the wavelength and period, remember that those numbers are multiplied by 2 pi as shown by your formula for a wave. Remember to divide those numbers by 2 pi in order to find the correct values for wavelength and period.

For the weight of the object: Wouldn't the weight of the object be equal to the tension in the string? Is there any way you can relate the tension to any of the known quantities?

Last edited: Oct 29, 2007
3. Oct 29, 2007

### azila

ok, i did what you said about the 2pi but even if I did that, I get the same velocity. My answer then is 23.8 s when I do velocity divided by length. And, yes by tension of the string do you mean like just multiplying the mass of the rope by 9.8m/s^2, I did that and that does not seem to work..thanks for the help though..any further help would be appreciated..

4. Oct 30, 2007

### saket

Tension in the rope would vary along its length (rope is not massless). Take the help of a free body diagram. As mass is uniform, it would be a linear variation -- (W + 1.24) at the top, W at the bottom.
Thus, T = W + 1.24*(x/L), where x is along the length of the rope and x = 0 at the bottomost point of the rope. {L is length of the rope.}
Now, accordingly, you can find speed variation of the wave along the length of the rope. And writing it as dx/dt, and integrating for x = 0 to x = L, you will get the time required.

5. Nov 6, 2007

### azila

I tried that, so I took the derivative of 1.24(1/L) and then integrated from 0 to 1.50 and did not work...any other explanation or is there something i am missing

6. Apr 28, 2008

### kjartan

Relevant equation: y(x,t) = Acos(kx - wt)
where k = 2pi/wavelength and w = 2pi*f

(d^2y(x,t)/dt^2)/(d^2y(x,t)/dx^2) = w^2/k^2 = v^2

So, v = |w|/|k| = (2730s^-1)/(172m^-1) = 15.9m/s

1.50m/15.9m/s = 0.0945s to go up the string

Next, m = w/g = 1.25N/9.81m/s^2 = 0.127kg
u = m/L = 0.127kg/1.50m = 0.0849kg/m
since v = (F/u)^(1/2), we have F = (v^2)u = 21.4N

Finally, to find how many waves are on the string at one time, we want to know L (length of the string) and lambda (wavelength)

w = 2pi*f = 2730s^-1 --> f = 434.5s^-1

lambda = v/f = 15.87m/s / 434.5s^-1 = 0.0365m
so, L/lambda = 1.50m/0.0365m/wave = 41.1 waves

Lastly, determine the equation for the waves traveling down the string.