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
