In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air is 330 m/s, the tuning fork frequency is :

Solution

Physics sir · 3 steps

The water rose 7.5 cm between two resonances, and that gap is exactly half a wavelength.

1. Air column shrinks. The heights are measured from the bottom, so raising the water from 17.0 cm to 24.5 cm shortens the air column by 24.5 - 17.0 = 7.5 cm. The end correction is the same for both resonances, so it cancels in this difference.
2. Successive resonances. In a tube closed at one end, the resonant lengths are λ/4, 3λ/4, 5λ/4, … Two neighbouring resonances therefore differ by λ/2.

   λ/2 = 7.5 cm ⇒ λ = 15 cm = 0.15 m

3. So. f = v/λ = 330/0.15 = 2200 Hz, which is option (a).

The idea to keep. For a resonance tube, the difference between two successive resonance lengths is always λ/2, and the end correction drops out. Use the difference, not either length on its own.

Answer from the NTA final answer key. Solution by PrepWiser.