# Vibrating membrane tension per unit length

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: 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.

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.

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 $$~T\,dx~$$

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