In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is:

Solution

Physics sir · 4 steps

The shorter pipe gives the higher note, so write both fundamentals with v/4L and subtract to get 15.

1. Closed pipe fundamental. A pipe closed at one end holds a quarter wavelength in its fundamental mode, so f = v/4L. The 100 cm pipe is shorter, so it gives the higher frequency.
2. Beat condition. The beat frequency is the difference of the two frequencies. Take L₁ = 1 m and L₂ = 1.2 m.

   v/4(1) - v/4(1.2) = 15 ⇒ v(1/4 - 1/4.8) = 15 ⇒ v/24 = 15

3. Check. With v = 360 m s⁻¹ the two pipes give 360/4 = 90 Hz and 360/4.8 = 75 Hz. Their difference is 15 beats per second, as required.
4. So. The speed of sound in the air column is v = 15 × 24 = 360 m s⁻¹, option (d).

The idea to keep. Two pipes beating against each other means you subtract their frequencies. Write each one in terms of v, and the difference equation hands you v directly.

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