Calculating Wave Pulse Travel Time for an 80ft Nylon Rope

AI Thread Summary
To calculate the wave pulse travel time along an 80 ft nylon rope under 3000 lb tension, the speed of the pulse is determined using the formula v = √(τ/μ), where τ is tension and μ is linear density. The linear density is calculated by dividing the mass (6 lb) by the length of the rope (80 ft). A user initially calculated the mass incorrectly, leading to an erroneous velocity and travel time of 2.27 seconds. The correct mass should be derived from weight divided by gravitational acceleration, which is 6 lb / 32.2 ft/s². This adjustment will yield the accurate travel time for the wave pulse.
daisyi
Messages
32
Reaction score
0
A nylon rope of length 80 ft. is under a tension of 3000 lb. The total weight of this rope is 6 lb. If a wave pulse starts at one end of this rope, how long does it take to reach the other end?
 
Physics news on Phys.org
Think of the mass (6lb) as being concentrated at the center of mass of the rope--at the geometric center in this case...

Cheers
Vivek
 
maverick280857 said:
Think of the mass (6lb) as being concentrated at the center of mass of the rope--at the geometric center in this case...
What good will that do?

The speed v of a pulse along a cord under tension \tau and with linear density \mu is given by:

v = \sqrt{\frac{\tau}{\mu}}

To evaluate this for v, you need to know the other two variables. You were already given the tension, now you need to find the linear density. The linear density is the mass divided by the length.

You can find the mass from the weight (which you were given). Divide the mass by the length, substitute that for \mu and the given tension for \tau, and you know how fast the pulse is travelling.

Solve displacement = velocity * time for time, plug in the value you just calculated for velocity and the length of the rope for displacement, and you're done.
 
Zorodius, your solution assumes that the mass of the string does not affect the overall motion. And I believe his problem was to figure out what role the mass has to do in that equation (which you answered). If I cut out the line mentioning the mass of the string and converted the problem into one where the tension and the linear density are given then it would take all the insight out of problem. Then it would just involve plugging the given values into v_{phase} = \sqrt{\frac{T}{\mu}}.

Of course, eventually we would all have used the formula for phase velocity that you posted, but my idea was to let him understand what he should do instead of giving him the solution rightaway. Besides, he didn't even send in his solution first (seeing which how we could have helped him better by helping him understand).

Anyways, you have correctly answered the problem and I have nothing against the solution :smile:



Cheers
Vivek
 
Last edited:
thanks for both of your help. The problem is that I already tried that (sorry I didn't put that there earlier, I was very frustrated at that point). But I am still coming up with the wrong answer. I keep getting 2.27 seconds...

so the mass = 6*32.2 = 193.2
and then to find u: 193.2/80 = 2.42
and then the velocity = sqrt(3000/2.42) = 35.2
so the time = 80/35.2 = 2.27 seconds.

grrr...what's going wrong here? :frown:
 
the problem is that mg=weight--->m=weight/g, and inthis case m=6/32.

I hope this helps
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top