Tension on a rope I was wondering how can one describe the tension force in a continuous way along a rope. We can look at a rope segment of length $dx$ and say the net tension force is $$T\sin\theta_{x+dx} - T\sin\theta_{x}$$ but that of course describes the tension force on a segment. I sometimes see that in texts they refer to the tension force at position $x$ as $-T\frac{\partial}{\partial x}\psi(x,t)$. How can one get from the first to the latter?

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*I refer to the tension along the vertical axis

 A: The proof can be found in any elementary text on waves. You can do it as follows: We follow the notation as in the figure.

Net force upward is
$$F_x(t)=T_2\sin\theta_2-T_1\sin\theta_2$$
We want to express $F_x(t)$ in terms of $\psi(z,t)$ and its space derivative $\partial \psi/\partial z$.
Now we make an approximation here, In the small-oscillation approximate, we neglect the increase in length of the segment, and we also approximate $\cos\theta$ by $1$. Thus we have $T\cos\theta=T_0$
$$F_x(t)=T_2\sin\theta_2-T_1\sin\theta_2$$
$$F_x(t)=T_2\cos\theta_2\tan\theta_2-T_1\cos\theta_2\tan\theta_2=T_0\tan\theta_2-T_0\tan\theta_2$$
$$F_x(t)=T_0\left(\frac{\partial \psi}{\partial z}\right)_2-T_0\left(\frac{\partial \psi}{\partial z}\right)_1$$
Now here we use taylor expansion and use the fact $\Delta z$ is small to approximate. If you work out you will get
$$\left(\frac{\partial \psi}{\partial z}\right)_2-\left(\frac{\partial \psi}{\partial z}\right)_1=\Delta \frac{\partial^2\psi(z,t)}{\partial z^2}$$
So that $$F_x(t)=T_0\Delta z\frac{\partial^2\psi(z,t)}{\partial z^2}$$

As a matter of fact the expression you have given is not quite correct.
