Two identical strings of length $L$ and linear mass density $\mu$ are under tension $T$. One string is plucked at $L/4$ and the other at $L/2$. What is the ratio of the energy of the second harmonic in the first string to the energy of the second harmonic in the second string?

Solution

1. The amplitude $A_n$ of the $n$-th harmonic for a string plucked at position $x_0$ is proportional to $\sin (n\pi x_0/L)$.
2. For the first string plucked at $x_0=L/4$, the amplitude of the second harmonic ($n=2$) is proportional to $$\begin{array}{rl}\sin (2\pi (L/4)/L)& =\sin (\pi /2)\\ & =1\end{array}$$.
3. For the second string plucked at $x_0=L/2$, the amplitude of the second harmonic ($n=2$) is proportional to $$\begin{array}{rl}\sin (2\pi (L/2)/L)& =\sin (\pi )\\ & =0\end{array}$$.
4. Since the second harmonic is not excited in the second string, the ratio of the energies is $1:0$.
5. Hence the answer is (A).