An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed and frequency is increased to f2 such that the resonance again occurs in nth harmonic. Choose the correct option. (A) n = 5, f2 = (5/4)f1 (B) n = 3, f2 = (3/4)f1 (C) n =5, f2 = (3/4)f1 (D) n =3, f2 = (5/4)f1

An open pipe is in resonance in 2nd harmonic with frequency f<sub>1</sub>. Now one end of the tube is closed and frequency is increased to f<sub>2</sub> such that the resonance again occurs in nth harmonic. Choose the correct option. (A) n = 5, f<sub>2</sub> = (5/4)ff<sub>1</sub> (B) n = 3, f<sub>2</sub> = (3/4)ff<sub>1</sub> (C) n =5, f<sub>2</sub> = (3/4)ff<sub>1</sub> (D) n =3, f<sub>2</sub> = (5/4)ff<sub>1</sub>

Answer:(A) n = 5, f2 = (5/4)f1

In open organ pipe

Fundamental harmonic of open organ pipe is given by:

  • ℓ = (λ1 / 2)
  • V1 = (V / λ1) = (V / 2ℓ)

As tube vibrates in second harmonic,

hence

f1 = 2V1 = (2V/2ℓ) = (V/ℓ)

In one end close pipe

If one end is close, it gives only odd harmonics & Fundamental frequency = (V / 4ℓ)

Other harmonics are (3V / 4ℓ), (5V / 4ℓ) etc.

One frequency starts increasing the first higher harmonic that is resonated is (3V/4ℓ)

If n = 3, then f2 = (3V/4ℓ) = (3/4)f1

However here is a snag.

The frequency is increased from (V/ℓ) hence (3/4).

f2 is not greater than f1.

∴ Answer is

  • n = 5
  • f2 = 5 × (V / 4ℓ) = (5/4)f1

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