A longitudinal mechanical wave propagates through a gas at a speed of 340\text{ m/s}. If a… — JAMB UTME

A longitudinal mechanical wave propagates through a gas at a speed of $340\text{ m/s}$ . If a particle in the gas completes $50$ full oscillations in $0.10\text{ s}$ , what is the distance between a compression and the immediate next rarefaction?

$0.34\text{ m}$ Answer
B
$0.68\text{ m}$
C
$1.70\text{ m}$
D
$1.36\text{ m}$

Answer

The distance between a compression and the adjacent rarefaction is

0.34\text{ m}

The frequency of the particle oscillation is

f = 50 / 0.10 = 500\text{ Hz}

. Using the wave equation

v = f\lambda

, the wavelength

\lambda = 340 / 500 = 0.68\text{ m}

. Because a compression and the consecutive rarefaction represent half of a full wave cycle, the distance between them is

\lambda / 2 = 0.34\text{ m}

Step-by-Step Solution

Calculate the frequency of oscillation of the wave

f = \frac{\text{Number of oscillations}}{\text{Time interval}} = \frac{50}{0.10\text{ s}} = 500\text{ Hz}

Frequency is defined as the number of complete vibrations per unit time.

Calculate the wavelength using the wave equation

\lambda = \frac{v}{f} = \frac{340\text{ m/s}}{500\text{ Hz}} = 0.68\text{ m}

The fundamental wave equation relates speed, frequency, and wavelength.

Determine the distance between consecutive compression and rarefaction

d = \frac{\lambda}{2} = \frac{0.68\text{ m}}{2} = 0.34\text{ m}

In a longitudinal wave, a compression and the nearest rarefaction are separated by half a wavelength.

Key Concept

Relation between frequency, wavelength, wave speed, and structural intervals in longitudinal waves

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