(i) A uniform stretched string of length $L$, mass per unit length $\rho$ and tension $T=\rho c^2$ is fixed at both ends. The motion of the string is resisted by the surrounding medium, the resistive force per unit length being $−2\mu \rho\dot y$ where $y(x,t)$ is the transverse displacement and $\dot y=\frac{\partial y}{\partial t}$. Show that the equation of motion of the string is $$\frac{\partial^2 y}{\partial t^2}+2\mu \frac{\partial y}{\partial t}=c^2\frac{\partial^2 y}{\partial x^2}$$ Find $y(x,t)$ if $\mu =\pi c /L$, $y(x,0)=d\sin(\pi x/L)$ and $\dot y(x,0)=0$

(ii) If an extra transverse force $F\sin(\pi x/L)\cos(\pi ct/L)$ per unit length acts on the string, find the resulting forced oscillation.

So the first part is quite straight forward, I got an answer in terms of fourier series, but I'm clueless when it comes to (ii). After some thought, I arrived at $\frac{\partial^2 y}{\partial t^2}+2\mu \frac{\partial y}{\partial t}=c^2 \frac{\partial^2 y}{\partial x^2}+\frac{F}{\rho}\sin(\frac{\pi x}{L})\cos(\frac{\pi c t}{L})$, but not sure how you can go from there, or if it's the right way.

  • 1
    $\begingroup$ For (ii) look at the derivation of the wave-equation on a string. Where you consider the transverse force (due to the string tension) you add the extra transverse force to get a modified wave-equation. Should be equivalent to something like adding $\frac{F}{\rho}\sin(\pi x/L)\cos(\pi ct/L)$ on the right hand side on the equation you got. $\endgroup$ – Winther Jul 19 '18 at 10:28

The fixed ends imply Dirichlet boundary conditions, i.e.

$$ y(0,t) = y(L,t) = 0 $$

Thus you can write the solution in series form as

$$ y(x,t) = \sum_{n=1}^\infty y_n(t)\sin\left(\frac{n\pi x}{L}\right) \tag{*} $$

Plugging the above form into the equation gives

$$ {y_n}'' + \frac{2\pi c}{L} {y_n}' + \frac{n^2c^2\pi^2}{L^2}y_n = 0 $$

The characteristic equation has roots

$$ r = -\frac{\pi c}{L} \pm i \frac{\pi c}{L}\sqrt{n^2-1} $$

The inital condition only includes the first eigenmode, so you can safely assume $y_n(t)= 0$ for $n > 1$. This leaves

$$ y_1(t) = (a_1 + b_1 t)e^{-\frac{\pi c }{L}t} $$

with $y_1(0) = d$ and ${y_1}'(0) = 0$

For the second part, you can start the same way, by assuming a series solution as in $(*)$. But this time, the equation for the first eigenmode contains the forcing term

$$ {y_1}'' + \frac{2\pi c}{L} {y_1}' + \frac{c^2\pi^2}{L^2}y_1 = \frac{F}{\rho}\cos\left(\frac{\pi ct}{L}\right) $$

you can solve this ODE using undetermined coefficients. Start by guessing a particular solution

$$ y_p = A\cos\left(\frac{\pi ct}{L}\right) + B\sin\left(\frac{\pi ct}{L}\right) $$

I trust that you can do the rest.



$$F\sin\left(\frac{\pi x}{L}\right)\cos\left(\frac{\pi c t}{L}\right) = \delta_{n1}\delta_{m1}F\sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi c t}{L}\right)$$ where $\delta_{ij}$ is the Kroneker delta defined as follows $$\delta_{ij}=\left\{\begin{matrix}1&\text{if }i=j\\0&\text{if }i\neq j\end{matrix}\right.$$

And note that $$\cos\left(\frac{m\pi c t}{L}\right)=1\;\;\forall m\;\;\;\text{when }t=0$$

I what to add that the equation you get by solving with the Fourier series method is quite ugly but somewhat standard and can be solved by means of Fourier transform


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