Waves Part 2

Concept:The fifth harmonic of a closed pipe is the $5^{\text{th}}$ overtone, which corresponds to $n=5$ in the formula $f = \frac{nv}{4L_c}$ (only odd $n$).The first harmonic (fundamental) of an open pipe is $f = \frac{v}{2L_o}$ (all $n$).Unison means equal frequencies.Formula:Closed pipe: $f_n = \frac{nv}{4L_c}$, $n=1,3,5,\dots$Open pipe: $f_n = \frac{nv}{2L_o}$, $n=1,2,3,\dots$Solution:Fifth harmonic of closed pipe: $f_c = \frac{5v}{4L_c}$First harmonic of open pipe: $f_o = \frac{v}{2L_o}$Set equal: $\frac{5v}{4L_c} = \frac{v}{2L_o}$Cancel $v$: $\frac{5}{4L_c} = \frac{1}{2L_o}$Cross‑multiply: $5 \cdot 2L_o = 4L_c \cdot 1$Simplify: $10L_o = 4L_c$Thus $\frac{L_c}{L_o} = \frac{10}{4} = \frac{5}{2}$Given $\frac{L_c}{L_o} = \frac{5}{x}$, so $\frac{5}{2} = \frac{5}{x}$Therefore $x = 2$Answer:$x = 2$ (Option B)