Q.A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz and 1152 Hz. Calculate the fundamental frequency.

[!TLDR] For a closed pipe, consecutive overtones differ by exactly $2n$: $896-640=256=1152-8 …

For a pipe closed at one end, the allowed frequencies are the ODD harmonics $n,3n,5n,7n,9n,\ldots$ (section 6.7.2/Long Answer 8), so any two CONSECUTIVE allowed frequencies always differ by exactly $2n$ (e.g. $5n-3n=2n$, or $7n-5n=2n$). Checking the given overtones: $896-640=256$ Hz, and $1152-896=256$ Hz -- the same common difference, confirming these are indeed three consecutive odd harmonics. So $2n=25 …

Use the fact that consecutive odd harmonics of a closed pipe always differ by exactly $2n$; find this common differenc …

Assuming the lowest given frequency (640 Hz) IS the fundamental itself, rather than correctly recognising these as higher (5th, 7th, 9th) o …