I have this exercise in my differential equations book…
If you pluck a violin string, and then finger the string, fixing it
precisely in the middle, the tone increases by one octave. In
mathematical terms this means that the frequency is doubled. Explain
why this happens.
Plucking the string without fingering should be represented as $$u_{tt} = c^2 u_{xx},\ u(0,\ t) = 0,\ u(L,\ t) = 0,\ u(x,\ 0) = f(x),\ u_t(x,\ 0) = 0$$ Plucking the string and then fingering should be represented as $$v_{tt} = c^2 v_{xx},\ v(0,\ t) = 0,\ v(\frac{L}{2},\ t) = 0 for\ t \geq t_1,\ v(L,\ t) = 0,\ v(x,\ 0) = f(x),\ v_t(x,\ 0) = 0$$
My game plan was to first solve for $u$, then find the first time at which $u = 0$ at $\frac{L}{2}$ to use as $t_1$, and finally solve for $v$ after $t_1$ is known. This means that the string is plucked from position $f(x)$ and allowed to follow its natural trajectory only until the first time the string crosses the $x$-axis at $\frac{L}{2}$, and is then pinned on the $x$-axis forever. However, the PDE in $v$ seems to have too many conditions for its order. Furthermore, the conditions of the PDE in $v$ are a superset of the conditions of the PDE in $u$, so I think the solutions to the PDE in $v$ should be a subset of the solutions to the PDE in $u$, even if it is somehow possible to apply this many conditions to the PDE in $v$. Are these PDEs possible to solve? Am I on the right track with this approach to the exercise?
Best Answer
The fundamental frequency of a string is
$$ f_0 = \frac{c}{2L} $$
if you half $L$, $L^{\rm (new)} = L / 2$ then, the new frequency is
$$ f_0^{\rm (new)} = \frac{c}{2 L^{\rm (new)}} = \frac{c}{2 L / 2} = 2\times \frac{c}{2L} = 2f_0 $$
that is: the fundamental frequency doubles