Question

A bat can hear sound at frequencies up to $120\,kHz$. Determine the wavelength of sound in the air at this frequency. Take the speed of sound in the air as $344\,m{s^{ - 1}}$.
(A) $1.87 \times {10^{ - 3}}\,m$
(B) $2.87 \times {10^{ - 3}}\,m$
(C) $3.87 \times {10^{ - 3}}\,m$
(D) $4.87 \times {10^{ - 3}}\,m$

Answer 1

Hint There is the common relation between the velocity of the wave, wavelength of the wave and the frequency of the wave. By using that relation and also by using the given information in that relation, the wavelength of the sound wave is determined.
Useful formula
The relation between the velocity of the wave, wavelength of the wave and the frequency of the wave is given by,
$\nu = \dfrac{v}{\lambda }$
Where, $\nu $ is the frequency of the sound wave, $v$ is the velocity of the sound wave and $\lambda $ is the wavelength of the sound wave.

Complete step by step solution
Given that,
The frequency of the sound wave is, $\nu = 120\,kHz = 120 \times {10^3}\,Hz$,
The speed of sound in the air is, $v = 344\,m{s^{ - 1}}$.
Now,
The relation between the velocity of the wave, wavelength of the wave and the frequency of the wave is given by,
$\nu = \dfrac{v}{\lambda }\,...................\left( 1 \right)$
By substituting the velocity of the sound wave and the frequency of the sound wave in the above equation (1), then the above equation (1) is written as,
$120 \times {10^3} = \dfrac{{344}}{\lambda }$
By rearranging the terms in the above equation, then the above equation is written as,
$\lambda = \dfrac{{344}}{{120 \times {{10}^3}}}$
On dividing the terms in the above equation, then the above equation is written as,
$\lambda = 2.8 \times {10^{ - 3}}\,m$
Thus, the above equation shows the wavelength of the sound wave for the given frequency and the velocity.

Hence, the option (B) is the correct answer.

Note The wavelength of the wave is directly proportional to the velocity of the wave. As the velocity of the wave increases, the wavelength of the wave also increases. The wavelength of the wave is inversely proportional to the frequency of the wave. As the frequency increases, the wavelength decreases.