Question

A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 Celsius, Find the frequency of the tuning fork.

Please answer and show work!!!

This is a Lab Report question btw. The lab report is called, "Resonance of air columns". Hope this sorta helps.

Answer 1

Given that, Room temperature is, T = 23^oC

Speed of sound at this temperature will be:

v = 331 + 0.61(T)

v = 331 + 0.61*(23)

v = 345 m/s

Wavelength will be:

λ = 2 (l₂ - l₁)

λ = 2*(34-12)

λ = 44 cm

λ = 0.44 m

Therefore, Frequency of the tuning fork (f) is given by:

λ*f = v

f = (v) / (λ)

f = (345 m/s) / (0.44 m)

f = 784.1 Hz ---------- (**Answer**)

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