Here is the problem I am woking on:

You are given a string of length L that is elastic. The function $v(x,t)$ measures the displacement of the string from the horizontal axis. At the left endpoint of the string ($x=0$) the string is fixed and $v(0,t)=0$. At the right endpoint of the string ($x=L$) the string is attached to a spring. The spring is only able to move up and down and it has elasticity k, linear density $\rho$ and tension force T (all constants). Moreover, the end of the spring not attached to the string is fixed at point $(L,h)$. We are given that the string has resistance force proportional to the displacement from the x axis. Overall the system is given by $\rho v_{tt} = T v_{xx} -r v$, with $v(0,t)=0$, and $T v_x(L,t)+k(v(L,t)-h)=0$.

How can we determine the equilibrium position of the string?

Secondly, if given the boundary condition $v(0,t)=sint$ how can we solve the BVP?

I'm entirely lost on where to start and what to do to solve this. From extensive google searching I think it may have something to do with the wave equation but I'm very lost on how to do any of this. Any suggestions, hints or explanations will be very appreciated.Even if you could just point me to some good reference material I would appreciate that.


1 Answer 1


At $x=L$, at rest, the forces acting on the movable hinge (positive downwards) will be
$T$, directed along the string;
$mg/2+k(v-h)$, directed along the vertical.


Normally, the end movement $v$ is taken to be much smaller than $L$.
Therefore $T$ is considered to be constant, so that its vertical projection would be approximated by $$ T_v = T{v \over L} $$

So that at equilibrium we shall have $$ mg/2 + k(v - h) + T_v = mg/2 + k(v - h) + T{v \over L} = 0 $$ which means $$ v(L,0) = {{ - mg/2 + kh} \over {k + {T \over L}}} $$ and this answers your first question.

Now consider that a little vertical starting displacement of the end at $x=L$, if much smaller than $L$, can be neglected, i.e. that $kh=mg/2$, which is to tell that the offset positioning of the spring is just to compensate the (half) weight of the string.
In conclusion at $x=L$ you may consider to have just a reaction (on the side of the string) of $-kv$.

The second step is that an elastic clamp as the above can be "bartered" with a little increase $\Delta L$ of the length $L$ and clamp the string to a fixed hinge there.
That's because near the end the displacement $v$ is even smaller than in the central part of string where it is already taken to be much less than $L$.
That allows to put $$ kv \approx T{{dv} \over {dx}} \approx T{v \over {\Delta L}}\quad \Rightarrow \quad \Delta L \approx {T \over k} $$ if the spring is stiff enough to give $\Delta L << L$. This allows as well to neglect the effect of the dynamic reaction of the final part of the spring.

If instead the spring is quite soft then we cannot adopt the approximation above: there will be reflections at the end .

Analog of the mechanical system above is the transmission along an electrical line of finite length, terminating on a capacitive load. This goes under the name of Telegrapher's Equation and you can find for that various sources for how to solve the problem.

--- clarifications ---

As a very concise reply to your comments:

a) The modelling of the tansverse vibration of a "normal" elastic string is derived as shown in this article as (using your notation) $$ \rho v_{\,t\,t} = Tv_{\,x\,x} $$ and this is analogous to the Telegrapher's equation for a lossless line.

b) The presence of the further term $-r v$ in the equation that you propose $$ \rho v_{\,t\,t} = Tv_{\,x\,x} - rv $$ corresponds to a string that lies on an elastic "bed" (with negligible mass and shear).
So the mechanical system is in any case lossless.

c) The $v(0,t) = \sin t$ is the exciting signal fed at $x=0$ in terms of an imposed displacement $v(t)$..
That means that the actual mechanical model underlying your mathematical model is the following


that is $$ \left\{ \matrix{ T_{\,n} (x) = T\,v_{\,x} \hfill \cr T_{\,n} (x + dx) - T_{\,n} (x) - r\,v\,dx = \rho dx\,v_{\,t\,t} \quad \Rightarrow \hfill \cr \Rightarrow \quad \rho v_{\,t\,t} = Tv_{\,x\,x} - r\,v \hfill \cr v(0,t) = U(t)\sin t \hfill \cr 0 = - T_{\,n} (L) - k\,v(L) = T\,v_{\,x} (L) + k\,v(L) \hfill \cr} \right. $$ where $U(t)$ denotes the Unit Step function.
Clearly this physical model is lossless.

d) For finding the steady state solution we adopt the method described in the Telegrapher's equation article and well known to electrical engineers,
and which essentially consists in taking the Fourier transform. We put $$ v(x,t) =\Re \left( {\hat V(x)\,e^{\,j\omega t} } \right) = V(x)e^{\,j\left( {\omega t + \alpha \left( x \right)} \right)} $$ where $\hat V(x)$ is in general complex, and replace in the system above.

Now that the system is in the correct shape (if I understood properly your post) can you proceed with the math part ?

  • $\begingroup$ @G Cab thank you very much for your response! Before I accept your answer officially I have a few questions about it because I still don't totally get it. Firstly, where does the given PDE come in? I may have missed it but I'm confused as to where that was used. Also, does the boundary condition $v(0,t)=sint$ get used? I may have missed that as well. $\endgroup$
    – k12345
    Commented Jan 22, 2021 at 22:47
  • $\begingroup$ @G Cab And lastly, I looked but the telegrapher's equation and the form I keep finding is $u_{tt} + \frac{1}{\tau} u_t = c^2 u_{xx}$ but it is missing a normal $u$ times some constant which is present in the PDE that was given in the problem. Is there a more general form of this equation that I'm not finding? I'm sorry about having so many questions. I hope you can help me understand some of this a little better. Thank you for your help! $\endgroup$
    – k12345
    Commented Jan 22, 2021 at 22:47
  • $\begingroup$ It's quite late for me now, and I will try to make my answer more clear tomorrow. But in the meantime pls. let me know which is your background in physics/engineering besides math $\endgroup$
    – G Cab
    Commented Jan 22, 2021 at 23:46
  • $\begingroup$ @G Cab Thank you very much! I'm sorry for my original response being so late. I don't have experience in physics/engineering at all. The extent of my physics comes from a high school physics class I took many years ago. I have an extensive math background though. $\endgroup$
    – k12345
    Commented Jan 22, 2021 at 23:55
  • $\begingroup$ @k12345: wish now I have been more clear, and since you have a good background in math, I leave to you to carry on $\endgroup$
    – G Cab
    Commented Jan 23, 2021 at 21:46

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