0
$\begingroup$

(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.

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Hint

$$F\sin\left(\frac{\pi x}{L}\right)\cos\left(\frac{\pi c t}{L}\right) = \delta_{n1}\delta_{m1}F\sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi c t}{L}\right)$$ 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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.