To solve this problem we will apply the concepts related to resonance. The velocity with which sound travels in any medium may be determined if the frequency and the wavelength are known. Since the pipe is closed at one end, that produces frequencies ratio 1:3:5, then
[tex]n_1 = \frac{V}{4L_1}[/tex]
Third resonance occurs at [tex]5n_2[/tex]
[tex]n_2 = \frac{V}{4L_2}[/tex]
At resonance [tex]5n_2=n_1[/tex]
[tex]5\frac{V}{4L_2} = \frac{V}{4L_1}[/tex]
[tex]L_2 = 5L_1[/tex]
[tex]L_2 = 5*0.2[/tex]
[tex]L_2 = 1m[/tex]
Therefore at 1m will the tuning fork again create a resonance condition
