Question

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 resonance again occurs in the nth harmonic. Choose the correct optio

Answer 1

Fundamental harmonic is one of the organ pipe is given by:  

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

Now the tube vibrates in second harmonic, isf1 = 2V1 = (2V / 2ℓ) = (V/ℓ)  

if one end is close,the old odd harmonics & Fundamental frequency                                        = (V / 4ℓ)  

And the other harmonics are (3V / 4ℓ), (5V / 4ℓ).

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

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

The frequency is increased from (V/ℓ) hence (3/4)   f1 is not greater than f1

{f1 = (V/ℓ)}.

Therefore f1 (5/4) is the answer

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