In a vibrating air column, the distance between a node and the adjacent antinode is 25 cm . If the speed of sound in air is 340 m s^ -1 , determine the frequency of vibration of the air column.

Correct answer: C. 340 Hz The distance between a node and the adjacent antinode is (λ)/(4) . (λ)/(4) = 25 cm = 0.25 m λ = 4 × 0.25 m = 1 m The frequency of vibration is given by f = (v)/(λ) . Substituting the given values: f = (340)/(1) = 340 Hz

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 a vibrating air column, the distance between a node and the...

In a vibrating air column, the distance between a node and the adjacent antinode is

25 \text{ cm}

. If the speed of sound in air is

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

, determine the frequency of vibration of the air column.

Options

A $1360 \text{ Hz}$
B $680 \text{ Hz}$
C $340 \text{ Hz}$
D $170 \text{ Hz}$

Correct answer

C. $340 \text{ Hz}$

Step-by-step solution

The distance between a node and the adjacent antinode is

\dfrac{\lambda}{4}

\dfrac{\lambda}{4} = 25 \text{ cm} = 0.25 \text{ m}

\lambda = 4 \times 0.25 \text{ m} = 1 \text{ m}

The frequency of vibration is given by

f = \dfrac{v}{\lambda}

.

Substituting the given values:

f = \dfrac{340}{1} = 340 \text{ Hz}

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