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Question. A wire horizontally over a bench has one end fixed and the other end holding some weight. The weight stretches the wire to produce some extension, say $\Delta x$. Someone puts a length marker somewhere on the wire except the two ends, and this marker indicates some extension, say $\delta x$. Does $\Delta x=\delta x$? If so, why does every point of the wire get extended by the same amount?


My attempt. If thinking about fixing one end of the wire and measuring how much the other end extends upon tension, I totally accept this notion of extension. However, how could every point on the wire get extended by the same amount? I am confused. Intuitively, I feel like every point of the wire contributes some extension which add up to the total extension. Any kind of help would be appreciated!

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    $\begingroup$ Try ss krotov problem 1.25 of mechanics here is the book pdf pdfdrive.com/… $\endgroup$ – Prateek Mourya Sep 10 '20 at 4:01
  • $\begingroup$ Hi, shelton. I was trying to find the e-book of ss krotov but failed. Do you know any link for that? Thanks a lot. @sheltonBenjamin $\endgroup$ – IncredibleSimon Sep 10 '20 at 4:08
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It is not true that every point of the wire gets extended by the same amount. Instead, every little length element gets stretched by the same amount, meaning you are right: those individual extensions all add up to the total extension.

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  • $\begingroup$ Thanks for your answer, niels. So does $\Delta x=\delta x$? The LHS seems to exceed the RHS if total extension is distributed among the wire. Thanks. @niels nielsen $\endgroup$ – IncredibleSimon Sep 10 '20 at 4:41
  • $\begingroup$ the right way to do it is by integrating, but I'm too damn old to remember how! $\endgroup$ – niels nielsen Sep 10 '20 at 5:26
  • $\begingroup$ when we take fixed end to be reference and the extension of a point is proportional to the distance of that point from reference point so rest you can work out $\endgroup$ – Prateek Mourya Sep 10 '20 at 5:39
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In the real world, if you pull a wire the part that will extend first is the one you hold in your hand. Then the tension will propagate further in the wire and eventually it will get uniform spacing. But not before a lot of oscillations happen. I know you are not exactly asking about the dynamics but it helps understanding what is really happening.

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  • $\begingroup$ Hmm... To be honest I couldn't visualize intuitively why it is the case. If I pull a spring say, isn't the tension gets spread along the spring uniformly? Thanks. @Exocytosis $\endgroup$ – IncredibleSimon Sep 10 '20 at 11:51
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    $\begingroup$ No, this is because energy travels at a finite speed. Imagine this: there are 4 boats on a lake, aligned and spaced uniformly. Now the first one starts to move forward. The second one wants to follow but it did realize the first boat was moving only after number one started its engine. So number two will move forward with a delay. Same with the other ones if they follow the boat just in front of them. There is a lag between the parts of a wire if you pull it. This is why you can break it if you pull it faster than the tension takes to dissipate into the rest of the wire. $\endgroup$ – Winston Sep 10 '20 at 12:04
  • $\begingroup$ That is indeed thorough. I understand now. So for this question, would $\Delta x=\delta x$ hold? I feel like the LHS should exceed the RHS, but my textbook assume this equality to hold when measuring the Young modulus of a copper wire. Thanks. @Exocytosis $\endgroup$ – IncredibleSimon Sep 10 '20 at 12:09
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    $\begingroup$ When? I assume after all is at rest. Then no, they are not equal. Delta x is the entire extension, and after everything settled down, it is spread uniformly assuming the wire material is uniform too. I don't know what LHS or RHS mean. $\endgroup$ – Winston Sep 10 '20 at 12:13
  • $\begingroup$ Haha, I am glad to find a flaw in my textbook. LHS and RHS mean left-hand side and right-hand side respectively. Thanks a lot, Exo. @Exocytosis $\endgroup$ – IncredibleSimon Sep 10 '20 at 12:21

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