A tuning fork is sounded above a resonating tube (one end closed),which resonates at a length of 0.3 m and again at 0.9 m.What is the frequency of the fork when the speed of sound is taken to be 340 m/s?

Question

Quizplus · Accepted Answer

The fundamental frequency of the resonating tube can be found using the formula: 
f = (n * v) / (2 * L)
where n=1 (since we are assuming the fundamental frequency), v=340 m/s (given), and L=0.3 m and 0.9 m (given for the two resonant lengths).
For L=0.3 m, 
f = (1 * 340) / (2 * 0.3) = 566.67 Hz
For L=0.9 m, 
f = (1 * 340) / (2 * 0.9) = 189.63 Hz
Since the tuning fork is causing the resonance, its frequency would be the same as the resonant frequency of the tube. Therefore, the answer is the resonant frequency closest to both 566.67 and 189.63 Hz. The only option that fits this criterion is B) 283 Hz.

A Tuning Fork Is Sounded Above a Resonating Tube (One