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I have been working on the problem of a vibrating membrane, with all the assumptions to make it ideal, but I still can't figure out why the tension per unit length multiply like that: Vibrating membrane if $T$ is the tension per unit length, why in the image on the right it is not multiplied by the length $\Delta x$, but by $\Delta y$ instead.

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2 Answers 2

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Think back to the 1D example of tension in a string, and look at the tensions at the ends of a small slice of length $\Delta x$. This is essentially the right-hand figure in your question, replacing $T\Delta y$ with the usual tension $T$.

Now, imagine stretching that string out along the $y$-axis a small distance $\Delta y$ to get an equivalent of your small membrane element. There is now a tension per unit length, and the total force is acting along this new length $\Delta y$. This means the tension force along the $x$-direction is given by $T\Delta y$. A similar argument holds for the tension force in the $y$-direction.

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enter image description here

Lets look at the equations for $~u(x,y)=u(x)$

Newton second low in transversal direction

$$dm\,\frac{dv}{dt}=F\sin(\beta)-F\sin(\alpha)\tag 1$$

with $\quad(~\alpha\ll~\beta\ll)$ $$dm=\mu\,ds\\ \tan(\alpha)=\frac{du}{dx}\quad,\alpha=\frac{du}{dx}\\ ds^2=dx^2+du^2\quad\Rightarrow\quad ds=\sqrt{1+\left(\frac{du}{dx}\right)^2}\,dx\approx dx\\ v=\frac{ds}{dt}=\frac{d x}{dt} $$ and $$\sin(\beta)-\sin(\alpha)\approx\alpha+d\alpha-\alpha=d\alpha=\frac{d^2u}{dx^2}\,dx$$

thus equation (1)

$$\mu\,dx\,\frac{d^2 x}{dt^2}=F\,\frac{d^2u(x)}{dx^2}\,dx$$

thus it should be $~T\,dx~$

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  • $\begingroup$ Sorry, I think you concluded wrong. The force should be $T\Delta y$ not $T\Delta x$ $\endgroup$
    – AmadoC
    Oct 22, 2022 at 21:21

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