Question

An open pipe is in resonance in 2^nd 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 the nth harmonic. Choose the correct option. 

(A) n = 3, f₃ = (3/4)f 

(B) n = 3, f₂ = (5/4)f₁ 

(C) n = 5, f₂ = (5/4)f 

(D) n = 5, f₂ = (3/4)f₁

Answer 1

The correct option (C) n = 5, f₂ = (5/4)f   

Explanation:

Fundamental harmonic of open organ pipe

ℓ = (λ₁/2) i.e. V₁ (V/λ₁) = (V/2ℓ).

As tube vibrates in second harmonic

hence

f₁ = 2V₁ (2V/2ℓ) = (V/ℓ) ----

if one end is closed it given only odd harmonics.

Fundamental frequency of closed organ

pipe = (V/4ℓ).

other harmonics are (3V/4ℓ), (5V/4ℓ) etc, once the frequency starts

increasing, 1^st higher harmonic that is resonated

= (3V/4ℓ)

if n = 3, then f₂ = (3V/4ℓ) = (3/4) f₁

but as frequency is increased from (V/L) here (3/4)f₁ is not greater than f₁ (= (v/ℓ)).

Hence (5/4) f₁ is answer as it is greater than f₁.

∴ n = 5 & f₂ = (5/4) f₁.