Test Index

Waves Part 2

Section: Physics

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Question : 7 of 54

Marks: +1, -0

In an open organ pipe

\nu_3

\nu_6

3^{\text{rd}}

6^{\text{th}}

harmonic frequencies, respectively. If

\nu_6-\nu_3=2200 \text{ Hz}

then length of the pipe is

\_\_\_\_

(Take velocity of sound in air is

330 \text{ m/s}

[22 Jan 2026 Shift 2]

200
225
275
250

Solution:

Concept:

For an open organ pipe, the

n^{\text{th}}

harmonic frequency is

\nu_n = \frac{n v}{2L}

, where

v

is the speed of sound and

L

is the pipe length.

Explanation:

The

3^{\text{rd}}

harmonic frequency:

\nu_3 = \frac{3v}{2L}

.

The

6^{\text{th}}

harmonic frequency:

\nu_6 = \frac{6v}{2L}

.

Given

\nu_6 - \nu_3 = 2200

Hz.

Substitute:

\frac{6v}{2L} - \frac{3v}{2L} = 2200

Simplify:

\frac{3v}{2L} = 2200

Plug

v = 330

m/s:

\frac{3 \times 330}{2L} = 2200

\frac{990}{2L} = 2200 \implies L = \frac{990}{2 \times 2200} = 0.225

m

Convert to mm:

L = 225

mm.

Answer:

225 mm (Option B).

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