In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning fork of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms⁻¹. The third resonance is observed when the air column has a length of _____________ cm.

Solution

Physics sir · 4 steps

Resonances in a closed tube are spaced λ/2 apart, so add one full λ to the first length.

1. Find the wavelength. λ = v/f = 336/400 = 0.84 m = 84 cm, so λ/4 = 21 cm.
2. Spot the end correction. The first resonance should be at λ/4 = 21 cm, but it is seen at 20 cm. The antinode sits slightly outside the open end, so the end correction is e = 1 cm.

   l₃ = 5λ/4 - e = l₁ + λ

3. Third resonance. The resonances come at λ/4, 3λ/4, 5λ/4, each minus the same e. So l₃ = 105 - 1 = 104 cm, which is the same as 20 + 84.
4. So. The third resonance is observed at an air column length of 104 cm.

The common slip: 100. Taking l₃ = 5 l₁ = 5 × 20 gives 100 cm. The first length is λ/4 - e, not λ/4, so multiplying it by 5 also multiplies the end correction.

The idea to keep. In a resonance tube the end correction is the same for every resonance. Use differences of lengths, lₙ₊₁ - lₙ = λ/2, and it cancels out.

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