a string with density $\rho$ and tension $T$ is bound at it's two ends at $x=0$ and $x=L$. there is a force acting on the string proportional to the velocity $F(x,t)= -2\gamma \rho \dot \psi(x,t)$ where $\psi(x,t)$ is the displacement of the string.

I'm having trouble understanding why the string equation is: $$\frac{\partial ^2 \psi}{\partial ^2t}-\frac{T}{\rho}\frac{\partial ^2 \psi}{\partial ^2 x}+2\gamma \frac{\partial \psi}{\partial t}=0 $$

It looks similar to the wave equation, just with damping added, but how can I prove this using only the information given?


A few remarks that might help…

(1) $\rho$ is the mass per unit length, not the density.

(2) The damping force quoted is per unit length of string.

(2) If you multiply the string equation you've quoted through by $\rho \delta x,$ you get the Newton's second law equation for length $\delta x$ of string.

(3) It should then be clear that the velocity-depemdent damping force has been correctly included.

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  • $\begingroup$ $\rho \delta x$ is mass. using newton's second law i got $\ddot \psi \rho \delta x=-2\gamma \rho \dot \psi+T\frac{\partial \psi}{\partial x}$. am I in the right direction? $\endgroup$ – segevp Dec 4 '18 at 20:12
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    $\begingroup$ First term on right needs multiplying by $\delta x$; see my remark (2). So does the last term. In last term you should have second derivative. It's not the local slope of the wire (first derivative) that gives the net force due to string tension on $\delta x$, but the rate of change of the slope (second derivative). See any textbook that derives the wave speed for a transverse wave on a string. $\endgroup$ – Philip Wood Dec 4 '18 at 20:51

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