Wave on string, bdry cond is mass on spring

[koopernikos]

Homework Statement

A semi-infinite string (constant tension T and density ρ) lies along the positive
x-axis. The motion of the string is described by the linear wave equation
\begin{equation}
h_{xx}-c^2h_{tt}=0
\end{equation}
with
c² ≡ T/ρ, where h(x,t) is the transverse displacement. Attached to the end of the string at
x = 0 is a mass m. Attached to the mass in the vertical direction is a spring with stiffness
constant k that has its other end attached to a rigid surface. The vertical displacement of
the mass m is designated by ξ (t) ≡ h(0,t). When the system is in equilibrium (i.e.,
when the spring is unextended and everything is at rest) ξ (t) ≡ 0. The radian frequency
of the free oscillator (when the string is not attached) is σ = k /m .
(a) By applying Newton's law to the motion of the mass at x = 0, find the boundary
condition on the string at x = 0. That is, find the equation of motion of the mass in
terms of ξ (t) . Ignore horizontal forces on the mass, and ignore gravity.
[HINT: the boundary condition should not have any terms showing h(0,t).]
(b) Prove that the total energy of the mechanical system (oscillator and string) is constant.
(c) Suppose that at t = 0 the initial conditions are

The Attempt at a Solution

a)
There are two forces (neglecting $\vec{g}$) acting on the mass: one due to the displacement (spring) and one stemming from the tension in the string.
\begin{align}
F_{net}
&=
\sum_i F_i\\
f_{net}
&=
\sum_i f_i\\
\ddot{\xi}
&=
-\sigma^2 \xi
+ c^2 h(0,t)_{xx}
\end{align}

b)
?
c)
?

...aaand that is about as much as i can come up with..

i played around with seperating variables a bit.. i don't even know what type of solution to expect.. something like
\begin{equation}
h(x,t)=e^{-At} (B \sin(kx-\omega t)+ C \cos(kx-\omega t))
\end{equation}
?
obviously
\begin{equation}
\sigma \neq \omega
\end{equation}

please help! I've been staring at the problem for 4 hours

 

[nilesh_pat]

and have gotten nothing. A:a) The force balance for the mass $m$ at $x=0$ is$$m\ddot{\xi}=-k\xi+T\,h_{x}(0,t).$$b) To prove the total energy is conserved, we need to calculate the total energy of the system which is given by $$E=\frac{1}{2}m\dot{\xi}^2+\frac{1}{2}k\xi^2+\frac{1}{2}\int_0^{\infty}T(h_x^2+h_t^2)\,dx.$$Now, differentiating with respect to time we get\begin{align*}\frac{dE}{dt}&=m\dot{\xi}\ddot{\xi}+k\xi\dot{\xi}+\int_0^{\infty}T(2h_xh_{xt}+2h_th_{tt})dx\\&=m\dot{\xi}(-k\xi+T\,h_{x}(0,t))+k\xi\dot{\xi}+T\,h_x(0,t)\,h_{xt}(0,t)+T\int_0^{\infty}(2h_th_{tt}-h_xh_{xx})dx\\&=m\dot{\xi}(-k\xi+T\,h_{x}(0,t))+k\xi\dot{\xi}+T\,h_x(0,t)\,h_{xt}(0,t)+T\int_0^{\infty}0dx\\&=m\dot{\xi}(-k\xi+T\,h_{x}(0,t))+k\xi\dot{\xi}+T\,h_x(0,t)\,h_{xt}(0,t)\\&=0.\end{align*}Note that the last integral is 0 since the wave equation holds everywhere, including $x=0$. Hence the total energy is conserved.c) We have the initial conditions:$$\xi(0)=\xi_