JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Waves & Sound

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The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe will be:

(A)60 cm
(B)45 cm
(C)30 cm
(D)15 cm✓ Correct

Answer: (D)

Solution

Fundamental frequency of a closed pipe of length

L_{1}

f_{1} = \frac{v}{4 L _{1}}

.

First overtone (second harmonic) of an open pipe of length

L_{2} = 60 cm

f_{2} = \frac{2 v}{2 L _{2}} = \frac{v}{L _{2}}

.

Setting

f_{1} = f_{2}

\frac{v}{4 L _{1}} = \frac{v}{L _{2}} ⟹ L_{2} = 4 L_{1} ⟹ L_{1} = \frac{60}{4} = 15 cm

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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