# Homework Help: Physics Kinematics Question

1. Sep 5, 2009

### skrapasor

1. The problem statement, all variables and given/known data
In a lecture demonstration, a 3.0-m-long vertical string with ten bolts tied to it at equal intervals is dropped from the ceiling of the lecture hall. The string falls on a tin plate, and the class hears the clink of each bolt as it hits the plate. The sounds will not occur at equal time intervals. Why? Will the time between clinks increase of decrease near the end of the fall? How could the bolts be tied so that the clinks occur at equal intervals?

initial velocity=v0=0 m/s
acceleration=gravity=a=9.80 m/s2
length of string=3.0 m

2. Relevant equations
x=x0+v0+(1/2)at2

3. The attempt at a solution
Well I know the answer, the time intervals will decrease. But why? It seems like common sense to me until I try to explain it by being specific and using equations. Assuming there is a bolt on each end of the string, then the interval between bolts is 0.10 m. But I don't actually think the length of the string is relevant, since we are told the intervals between bolts are equal distances. The length could be anything assuming it will fit in the lecture hall, so a more general answer is preferred. I'm in AP Physics, and I'm trying to give an answer that would be a good score on the AP test that shows complete understanding. The solution that was given in the book didn't even make sense. I'm not going to post it because I don't want it to affect your answers. My teacher hasn't even assigned homework so don't think I'm trying to be lazy, I just really can't get past this question.

Here is my attempt: "Since v2=v02+2a(x-x0), with greater displacement (x), velocity at the point 0.10 m above the plate will be greater and thus with each bolt the time elapsed will be shorter due to x=x0+v0t+(1/2)at2, so if x stays the same as does a, when v0 increases (velocity of the bolt at 0.10 m above the ground), t decreases. Since time elapsed is a function of the square root of distance over acceleration because of x=x0+v0+(1/2)at2 solved for t, since acceleration remains constant, distance must increase by a factor of the square of the multiple of the time interval between the first two clinks." I really would like to say what I mean concisely and more accurately.

2. Sep 5, 2009

### loveequation

First note that at each instant during its fall, the string and all the attached weights will be dropping at the same speed a*t, i.e., as if the whole system were a rigid body. This is because all of them have been falling for the same time. This is the "time fixed" picture.

But now consider the situation at a fixed height "y" and imagine the weights successively going thru this height. The ones arriving later will have been traveling for a longer time and hence will be going faster. Hence the clinks will occur faster.

To get the clinks to occur in equal time intervals you need to space the weights further apart the higher you go up the string. This must be done quadratically. That is, if the length spacing between the first weight and the second is unity, then the next spacing should be four, the next nine, etc.

I will explain why shortly after I run an errand.

3. Sep 5, 2009

### loveequation

Let y = 0, y1, y2, etc. be the locations of the weights up the string measured from its bottom. Now, intuitively, the height of the floor/tin from the bottom of the string (y = 0) should be immaterial. If you get it to work for one height, it will work for all heights. So you might as well let the floor/tin be at y = 0. The first weight will hit the tin at t = 0, the next at t1 = (2 y1/g)^(1/2) and the Nth at tN = (2 yN/g)^(1/2). We want tN = N*t1. This gives yN = N^2 t1.

If you want to prove that the height of the string above the floor/tin is immaterial, just add H to y1 and yN in the above discussion and notice that its drops out.

4. Sep 5, 2009

### skrapasor

Thanks! That makes it much more clear to me now.