A long tube contains air at a pressure of 1.00 atm and a temperature of 77.0ºC. The tube is open at one end closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of 500 Hz. Resonance is produced when the piston is at distance 18.0, 55.5, and 93.0 cm from the open end.
Text Solution
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The second distance is midway between the first and third, and if there are no other distances for which resonance occurs, the difference between the first and third positions is the wavelength λ = 0.750 m. (This would give the first distance as λ/4 = 18.75 cm, but at the end of the pipe, where the air is not longer constrained to move along the tube axis, the pressure node and displacement antinode will not coincide exactly with the end). The speed of sound in the air is then v = fλ = (500 Hz)(0.750 m) = 375 m/s.
Solving Eq. v = 
(speed of sound in an ideal gas) for γ,
γ= 
= 
= 1.39.
Since the first resonance should occur at t/4 = 0.875 m but actually occurs at 0.18 m, the difference is 0.0075 m.
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