The second overtone of an open pipe is in resonance with the first overtone of a closed pipe of length 2 m. Length of the open pipe is

Options

(a) 4 m
(b) 2 m
(c) 8 m
(d) 1 m

Correct Answer:

4 m

Explanation:

Frequency of second overtone of an open is ʋ = 3v / 2l₀ Frequency of first overtone of a closed pipe is ʋ’ = 3v / 4lc Given : ʋ = ʋ’ ⇒ 3v / 2l₀ = 3v / 4lc l₀ / lc = 2 ⇒ l₀ = 2lc = 2 × 2 = 4 m.