In an open organ pipe, two successive resonance frequencies are 1620 Hz and 2268 Hz . Determine the length of the tube. The speed of sound in air is 324 m s^ -1 .

Correct answer: D. 25 cm The difference between two successive resonance frequencies is: Δ f = 2268 - 1620 = 648 Hz For an organ pipe, the difference between successive resonance frequencies is (v)/(2L) . (v)/(2L) = 648 (324)/(2L) = 648 2L = (324)/(648) = (1)/(2) L = (1)/(4) m = 25 cm

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Sound Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Sound Waves

In an open organ pipe, two successive resonance frequencies are...

In an open organ pipe, two successive resonance frequencies are

1620 \text{ Hz}

and

2268 \text{ Hz}

. Determine the length of the tube. The speed of sound in air is

324 \text{ m s}^{-1}

Options

A $20 \text{ cm}$
B $50 \text{ cm}$
C $12.5 \text{ cm}$
D $25 \text{ cm}$

Correct answer

D. $25 \text{ cm}$

Step-by-step solution

The difference between two successive resonance frequencies is:

\Delta f = 2268 - 1620 = 648 \text{ Hz}

For an organ pipe, the difference between successive resonance frequencies is

\dfrac{v}{2L}

\dfrac{v}{2L} = 648

\dfrac{324}{2L} = 648

2L = \dfrac{324}{648} = \dfrac{1}{2}

L = \dfrac{1}{4} \text{ m} = 25 \text{ cm}

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