The Physics of Tension in a Phone Cord

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The discussion centers on calculating the tension in a phone cord using its mass, length, and the time taken for a wave pulse to travel along it. Participants express confusion over the correct formula and calculations, with one user attempting to derive tension using the equation T = v^2(m/L). Despite repeated attempts, the calculations yield inconsistent results, leading to frustration. There’s a consensus that careful attention to detail and step-by-step calculations are essential to avoid errors. The problem is also critiqued for its unrealistic parameters, suggesting it may not accurately reflect real-life scenarios.
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Question:

A phone cord is 3.75 m long. The cord has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.915 s. What is the tension in the cord?

Maybe I'm using the wrong formula or my math is off. Help?

M = .200kg
L = 3.75m
P = .915s

U=M/L= .200/3.75 = .0533

C=(3.75)(4)/.915= 16.3934

(16.3934)^2 (.0533) = 143.2

But, 143.2 isn't right.

Thanks for help. I really hate it when I think I know what I'm doing, but I don't get the answer. It's really annoying.
 
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Let's check a couple things. What equation are you using to relate the mass per unit length, the propagation velocity, and the tension? And in your last line of calculations, be sure to include units to serve as a check that you haven't left anything out.


EDIT -- Pretty funny, I just saw your PF handle. Well, welcome to PF anyway! :smile:
 
I can't find the equation I used for what's shown above, for some reason.

I did find this one online: v = (T/(m/L))^0.5

and when I re-arranged it to find T, I got v^2(m)(L)=T which, when I plugged in the numbers, came out to .0021, which doesn't seem right.

I don't know what I'm doing. I don't get physics at all... even with a tutor :(
 
I found a similar problem online:

A phone cord is 4.20 m long. The cord has a mass of 0.180 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.745 s. What is the tension in the cord? ___ N

Solution 23660

Mass: m = 0.180 kg
Length: L = 4.20 m
u = m / L = 0.180 / 4.20 = 3/70
c = (4.20 * 4) / 0.745 = 3360/149
T = (3360/149)^2 * (3/70) = approximately 21.8
The tension is 21.8 N.


I don't know what formula they used, but that's the same thing I did at first. I just plugged in my numbers again and I'm still getting the wrong answer.
 
PhysicsHatesMe01 said:
...
I did find this one online: v = (T/(m/L))^0.5

and when I re-arranged it to find T, I got v^2(m)(L)=T ...

You need to take it slow. Work on the paper. That is do not skip steps. Keep it simple - do not try to leave out obvious steps. That way you tend to make errors. The formula conversion to T above is not corrcect.
 
Keeping(m/L) together, I now got:

V^2*(M/L)=T

and I get 16.39^2/.0533=5042 and that seems too high.

I know I should square v because anything to the 1/2 power = a sq. root...
 
I get the same answer. It is difficult to put the answer into perspective. Hanging 500 kg (6 persons!) from a 200 g string (quite a thick/heavy string for its length)? Causing a disturbance to travel at only 16 m/s (seems a bit slow for the given tension)? At what rate will it vibrate? This seems like an artificial problem to me - it probably does not represents something in real life.
 
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