Why is this a Mechanics problem?

If the first harmonic has frequency $f$ , the third harmonic has frequency $3f$ . This is just intonation.

If the fundamental has frequency $f$ , an octave and a fifth up has frequency $2^{19/12}f$ . This is equal temperament.

These tuning systems are so close that $2^{19/12}$ and $3$ have an absolute difference of about $0.003$ .

Well $3-2^{n/12}\approx 0\implies 2^n\approx 3^{12}=531441$ and $2^{19}=524288$ so $\fbox{19}$ is the best approximation.

$\displaystyle{\\ ({ 2 })^{ \cfrac { n }{ 12 } }\approx 3\\ n\approx 12\times \cfrac { \ln { 3 } }{ \ln { 2 } } \\ n\approx 12\times \cfrac { 1.1 }{ 0.7 } =12\times 1.57=(10+2)\times 1.57=15.7+3.14=18.84\\ \boxed { n\approx 19 } }$

Note : I don't use calculator Since value of log(2) and log(3) is very standard data ... which have to learn while solving questions of chemistry....

So from my views It is chemistry question.... :)