Vibrating String — Question 8

Question 8 – Solution

Strategy. The midpoint mass changes the dynamic matching condition, but modes with a node there do not move the added mass.

Step 1: Separate and use symmetry. On each half, $\phi''+k^2\phi=0$, $k=\omega/c$, and the mass equation becomes $$T(\phi'_+-\phi'_-)+M\omega^2\phi(L/2)=0.$$ Antisymmetry gives $\phi(L/2)=0$ and matching slopes. The fixed endpoints then yield $$\boxed{\phi(x)=A\sin(2m\pi x/L),\qquad \omega=2m\pi c/L,\quad m=1,2,\ldots.}$$ These frequencies are independent of $M$.

Step 2: Derive and locate the symmetric frequencies. A symmetric shape has $\phi=A\sin(kx)$ on the left and $\phi=A\sin(k(L-x))$ on the right. Its slope jump is $-2Ak\cos z$ and midpoint value is $A\sin z$. Substitution gives $$2Tk\cos z=M c^2k^2\sin z,\qquad \boxed{\cot z=\alpha z,\quad \alpha=\frac{M}{\mu L}.}$$ On $(j\pi,j\pi+\pi/2)$, $\cot z-\alpha z$ decreases strictly from positive infinity to a negative value. It has exactly one root. On the intervening intervals the cotangent is negative, so no positive root occurs there. Values $\sin z=0$ do not solve the original matching equation. There is no nonzero static mode with both endpoints fixed.

Step 3: Interpret the lowest root. For the root $z_0\in(0,\pi/2)$, $\alpha\downarrow 0$ gives $z_0\to\pi/2$, recovering $\omega_0\to\pi c/L$. As $\alpha\to\infty$, $z_0\to 0$ and $z\cot z\to 1$ gives $\alpha z_0^2\to 1$. Thus $$\boxed{\omega_0\sim 2\sqrt{\frac{T}{ML}}\quad(M\to\infty).}$$ A small midpoint deflection $d$ gives each half-string slope magnitude $2d/L$; their restoring forces add to $4Td/L$. The effective stiffness is $4T/L$, consistent with the displayed mass-spring frequency.

Step 4: Include the concentrated kinetic energy. For piecewise smooth motion, the energy is $$E=\frac 12\int_0^L(\mu u_t^2+Tu_x^2)\,dx+\frac M2u_t(L/2,t)^2.$$ Integration by parts on each half leaves $T u_t(L/2)(u_x^- -u_x^+)$; the fixed-end contributions vanish. The point-mass term differentiates to $M u_t(L/2)u_{tt}(L/2)=T u_t(L/2)(u_x^+-u_x^-)$. They cancel, so $\boxed{E'=0}$. Omitting the mass term would falsely suggest loss or gain of energy at the interface.