I always assumed tension is uniform in a wave but i saw a question in which we had to form an equation for tension at different position and time. The answer given was $$ T \sqrt{ 1+\left(\frac{\partial y}{\partial x}\right)^2 }.$$ Where $T$ I assume is the uniform tension which we usually talk about. Please help .Sorry if there is any problem with my query.
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I think you are confusing the tension with the potential energy of a stretched string. If a string under tension $T$ is slighly stretched by a distance $dl$ the work done to stretch it is $Tdl$. This assumes that the change in length is small so that the change in $T$ due to the stretch is negligible (i.e of second order in $dl$). If the string, originally lying along the $x$, axis from $0$ to $L$, is bent so that its profile becomes $y(x)$ the change in length is $$ \delta L= \int_0^L \sqrt{1+\left(\frac {\partial y}{\partial x}\right)^2} dx- L $$ so the potential energy stored in the string is $$ T \delta L= \int_0^L T \left(\sqrt{1+\left(\frac {\partial y}{\partial x}\right)^2}-1\right) dx\\ \approx \int_0^L \frac 12 T \left(\frac {\partial y}{\partial x}\right)^2dx. $$ There are cases where we need to consider the change in $T$, but these require knowledge of the Young's modulus of the string, and this does not seem to be what you are asking about.
Note added. I just saw that I answered essentially the same quastion here. I am getting old. I have no recollection of this previous answer....
mike stone
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If you multiply your expression by dx/dx, it can be expresed as T(ds/dx) where the (ds) is then length of a short segment after it is tilted. This implies that the tension in each segment is proportional to the length of the segment (which is not true).
R.W. Bird
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