2
$\begingroup$

In case of a home experiment about string vibration under the boundary condition $$y(l,t)=y(0,t)=0$$ Where $y=$ the displacement of the string at spatial co-ordinate $x$ and at time $t$, I observed that under any kind of arbitrary initial condition as the wave damps out with time ultimately it vibrates in the first normal mode for some time after all other modes having been damped out before stopping.

Physically it can be argued that as the frequency corresponding to higher normal modes are higher the velocity of the different portions of the string due to the higher normal modes will also be higher and due to high velocity it will be damped very fast as the damping in real life is proportional to velocity. But I am failing to mathematically model this problem to have a quantitative idea about the damping of the individual normal modes. Now,how can I mathematically model this damping?

$\endgroup$

3 Answers 3

1
$\begingroup$

The free vibration of a string can be modelled with the partial differential equation:

$$T\frac{\partial^2 y(x,t)}{\partial x^2}-\rho \frac{\partial^2 y(x,t)}{\partial t^2} = 0$$

Where $T$ is the axial tension on the string, $\rho$ is the mass density, $y$ is your transverse displacement, and $x$ is the position along the string length. The solution of this equation can be obtained using, for instance, the method of separation of vibrations ($y(x,t) = Y(x)\exp(\omega t)$), and later applying your boundary conditions.

Damping can be very difficult to predict, but a viscous type damping model can be incorporated in this model as

$$T\frac{\partial^2 y(x,t)}{\partial x^2} + \beta\frac{\partial y(x,t)}{\partial t}-\rho \frac{\partial^2 y(x,t)}{\partial t^2} = 0$$.

Where $\beta$ is a viscous damping coefficient, and since the damping forces are proportional to the velocity, the higher is the frequency, the higher are the damping forces. So each mode of vibration is affected differently by damping.

$\endgroup$
-1
$\begingroup$

Try the following: 1. compare the test between vertical and horizontal settings, to see the gravitation effect, which is most likely in play in your test; 2. plug in different initial modes: first mode, plug the midpoint; second mode, plug two points with an offset but in opposite direction; so on; 3. change the rigidity of your boundary condition: your BC may be not rigid enough

$\endgroup$
-1
$\begingroup$

First, the question contains an error because it assumes the initial condition of the string is determined by an “arbitrary initial condition” but the initial condition is the string length and tension, not the position of the string when oscillation begins.

The second answer absurdly implies that the string action is not independent of its coordinate system and how the string is embedded in Euclidean space. The string is a natural invariant with an interesting geometric interpretation that Stack Exchange doesn’t seem to understand.

For example, there two fatal flaws in first answer that no one has noticed in the last five years:

  1. The correct equation is the time-independent form of the hamiltonian function H(p, q) = h, not H(p, q, t) since the string is a standing wave dH/dt = 0. The time-independent hamiltonian is far better because it means the lines of motion are precisely the reduced action extremals of the integral ∫_s▒〖p dq〗 of the class of curves s lying on the surface of the manifold (See Arnold “Mathematic Methods of Classical Mechanics” 2010, pg 245 and Frankel “The Geometry of Physics” 2012 pg 147 for the complete equations).
    The differential equation you have written cannot described a line of motion that is on the string manifold.
  2. Your equation of motion assumes the force acting upon the string is proportional to transverse displacement. But that is true only if the string elongates as it bends, which it clearly does not the case. If the string cannot elongate (cannot move unless acted upon by external force), the Fourier analysis cannot be used. The increment ∆x = 0 is the boundary condition on x that makes the string curve a cosh x function. Clearly, it cannot be true that when a point of the string q_i reaches its outer limit (where potential energy would be greatest if the string stretches), then the velocity of the string goes to zero and the string returns to the center of motion where velocity is greatest. The force vector you think is induced by displacement points in the direction of the center of motion, therefore the string is confined, it seems, to a plane through the y-axis at an arbitrary angle.

Clearly this is nonsense. The string orbit lies in a plane that is a disc passing through the point and perpendicular to the string axis since ∆x = 0. Then the Liouville integral integrates the phase space like a union form stack of coins. Then under uniform motion the velocity of a point on the string is constant when it reaches boundary because the velocity is a constant of motion and not a function of t. Time is just a dummy variable on the string because the string is itself a uniform, invariant measure of time and distance. The string manifold is not a sine wave. There seems to be a failure of physicists to see the string manifold is symplectic. Is there anyone out there who can prove the curvature of the string is not constant?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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