Solution:
Frequency \(f \propto \frac{1}{L}\), so \(\frac {f_1}{f_2} = \frac{26}{25}\). Since \(\f_1 -\f_2 = 10\), we have \(26x - 25x = 10 \Rightarrow x = 10\). Thus \(f_1 = 260\text{ Hz}\) and \(f_2 = 250\text{ Hz}\).
Two closed pipes produce \(10\text{ beats per second}\) when emitting their fundamental modes. If their lengths are in ratio of \(25 : 26\). Then their fundamental frequencies in Hz are
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