A tuning fork with a frequency of $800 ext{ Hz}$ produces resonance in a resonance column tube with the uppe

Question

A tuning fork with a frequency of $800 	ext{ Hz}$ produces resonance in a resonance column tube with the upper end open and the lower end closed by the water surface. Successive resonances are observed at lengths of $9.75 	ext{ cm}$, $31.25 	ext{ cm}$, and $52.75 	ext{ cm}$. The speed of the sound in the air is:

Accepted Answer

In a resonance column tube (which acts as a closed organ pipe), the difference between two successive resonance lengths is equal to half the wavelength ($\lambda/2$).
Given successive resonance lengths:
$l_1 = 9.75 	ext{ cm}$
$l_2 = 31.25 	ext{ cm}$
$l_3 = 52.75 	ext{ cm}$
We can find the wavelength $\lambda$:
$\frac{\lambda}{2} = l_2 - l_1 = 31.25 	ext{ cm} - 9.75 	ext{ cm} = 21.5 	ext{ cm}$
(Also verified by $l_3 - l_2 = 52.75 - 31.25 = 21.5 	ext{ cm}$)
$\lambda = 2 	imes 21.5 	ext{ cm} = 43.0 	ext{ cm} = 0.43 	ext{ m}$
Given the frequency of the tuning fork, $f = 800 	ext{ Hz}$, the speed of sound $v$ is:
$$v = f\lambda = 800 	ext{ Hz} 	imes 0.43 	ext{ m} = 344 	ext{ m/s}$$

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