For a pipe closed at one end and open at the other end, the resonance frequency is given by:

$\overline{){\mathbf{f}}{\mathbf{=}}\frac{\mathbf{n}\mathbf{v}}{\mathbf{4}\mathbf{L}}}$

Wavelength:

$\overline{){\mathbf{\lambda}}{\mathbf{=}}\frac{\mathbf{v}}{\mathbf{f}}}$

L_{1} is greater than L_{2}

Therefore, the resonance frequency of pipe 1 is:

Using the two measured pipe lengths (L_{1} = 66 cm and L_{2} = 40 cm), work out the wavelength of the sound wave. use this to determine the mode numbers and speeds of sound that the two lengths correspond to. You can assume that L_{1} and L_{2} represent neighboring resonances (i.e, n and n+2). the pipes are open on one end and closed on the other. frequency of tuning fork is 384 Hz.

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