A pipe of length 1.0 m is used in two ways, first closed at one end and then open at both ends. The speed of sound is 340 m s^-1; neglect end…

A closed pipe has a node at the closed end and an antinode at the open end, so only odd harmonics occur: $f_n=(2n+1)\dfrac{v}{4L}$ for the $n$-th overtone, with $\dfrac{v}{4L}=85\ \mathrm{Hz}$. (P) The fundamental is $85\ \mathrm{Hz}$, so P→1. (Q) The third overtone is the $7$th harmonic, $7\times85=595\ \mathrm{Hz}$, so Q→5 (the $255\ \mathrm{Hz}$ value is the third harmonic, i.e. the first overtone). An open pipe has all harmonics $f_n=n\dfrac{v}{2L}$ with $\dfrac{v}{2L}=170\ \mathrm{Hz}$. (R) The fundamental is $170\ \mathrm{Hz}$, so R→2. (S) The second overtone is the third harmonic, $3\times170=510\ \mathrm{Hz}$, so S→4. The matching is P→1, Q→5, R→2, S→4.