Is my solution for the Bungee Tension Problem correct?

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    Bungee Tension
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Homework Help Overview

The discussion revolves around a bungee tension problem involving forces, tension, and wave velocity. Participants are analyzing the forces acting on a mass during bungee jumping and exploring the relationship between tension in the cord and the displacement caused by the mass.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss summing forces at different points in the motion and derive expressions for tension and displacement. There is an exploration of using energy conservation as an alternative method. Questions arise regarding the assumptions made about the forces and the relationship between tension and displacement.

Discussion Status

Some participants have provided alternative approaches and questioned the assumptions made in the original calculations. There is ongoing exploration of the relationships between tension, displacement, and wave velocity, with no explicit consensus reached yet.

Contextual Notes

Participants note the importance of distinguishing between the mass of the block and the mass of the cord, as well as the need to justify assumptions made about the forces at different points in the motion.

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Homework Statement
A block with mass M is hanging from a rubber cord, supported such that the cord is unstretched at length L0. The mass of the cord is m which is much smaller than M. The cord has spring constant K. If the block is released: Find the tension, length of cord and speed of wave at the lowest point.
Relevant Equations
V=sq.rt T/K
Sum of the forces
Conservation of energy
Hello again,
I did a sum of the forces at the top and bottom of the motion and got:
Top...mg=ma
Bottom...T-mg=ma
Got T=2mg when I substitute the mg in for ma at bottom.Then I assumed kx=T=2mg so x(length of cord) would equal 2mg/k
Then V of wave would be ...sq.rt2mg/k
Is it that simple? Feel like I'm missing something. Thank you for your help
 
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norcal36 said:
Homework Equations:: V=sq.rt T/K
Check your notes. I believe that K here should be replaced by the mass per unit length of the cord.
(You can use the editing toolbar to get symbols such as the square root symbol √ )

I did a sum of the forces at the top and bottom of the motion and got:
Top...mg=ma
Bottom...T-mg=ma
Got T=2mg when I substitute the mg in for ma at bottom.
It appears to me that to get this result you have assumed that the value of ma at the top is the same as the value of ma at the bottom. This might be true. But it needs to be justified.

Another approach for finding the distance x that the mass M falls is to use energy concepts.
 
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You are right the equation I used is for another situation. However if I use the conservation of energy I get mgx=1/2kx^2...mg=1/2kx...kx=2mg...kx=T=2mg. Again I am assuming tension will be the equivalent to kx. The x would be the displacement of the "spring" which equals length. However I am still stuck on the velocity of the wave. thanks for your time again.
 
norcal36 said:
You are right the equation I used is for another situation. However if I use the conservation of energy I get mgx=1/2kx^2...mg=1/2kx...kx=2mg...kx=T=2mg. Again I am assuming tension will be the equivalent to kx.
OK.
The x would be the displacement of the "spring" which equals length.
x is how much the cord is stretched from its natural length. So, how would you express the total length of the cord when M is at the bottom?

Then, can you find an expression for the mass per unit length of the cord when it is at the bottom?
 
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I would express the total length by L0+x...?
Then with that I could use V=√T/m/L equation which then would give me V=√2g(L0+x)...This sound correct?
 
norcal36 said:
I would express the total length by L0+x...?
Yes
Then with that I could use V=√T/m/L equation which then would give me V=√2g(L0+x)...This sound correct?
Be sure to distinguish between the mass of the block and the mass of the cord.
 
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