Question

A bat emits ultrasonic sound of frequency \[1000\,{\text{kHz}}\] in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound. (b) the transmitted sound? Speed of sound in air is \[340\,{\text{m}}{{\text{s}}^{ - {\text{1}}}}\] and in water is \[1486\,{\text{m}}{{\text{s}}^{ - {\text{1}}}}\] .

Answer 1

Hint: First, convert the frequency into regular units. Find the formula which relates velocity, frequency and wavelength; substitute the values and manipulate accordingly.

Complete step by step solution:
(a) In the given problem, the frequency of the sound which is ultrasonic in nature is \[1000\,{\text{kHz}}\] .
\[
 n = 1000\,{\text{kHz}} \\
 = 1000 \times {10^3}\,{\text{Hz}} \\
 = {10^6}\,{\text{Hz}} \\
\]
Speed of the sound \[\left( {{v_a}} \right)\] in air is given as \[340\,{\text{m}}{{\text{s}}^{ - {\text{1}}}}\] .
The wavelength of the reflected sound is related to the frequency by the following expression:
\[{\lambda _r} = \dfrac{{{v_a}}}{n}\] …… (1)
Where,
\[{\lambda _r}\] indicates the wavelength of the reflected sound.
\[{v_a}\] indicates the velocity of sound in air.
\[n\] indicates the frequency of the emitted sound.

Substituting the values of frequency and velocity of the sound wave in equation (1), we get:
\[
 {\lambda _r} = \dfrac{{{v_a}}}{n} \\
 = \dfrac{{340\,{\text{m}}{{\text{s}}^{ - 1}}}}{{{{10}^6}\,{\text{Hz}}}} \\
 = 3.4 \times {10^{ - 4}}\,{\text{m}} \\
\]
Hence, the wavelength of the reflected sound is found to be \[3.4 \times {10^{ - 4}}\,{\text{m}}\] .

(b) In the given problem, the frequency of the sound which is ultrasonic in nature is \[1000\,{\text{kHz}}\] .
\[
 n = 1000\,{\text{kHz}} \\
 = 1000 \times {10^3}\,{\text{Hz}} \\
 = {10^6}\,{\text{Hz}} \\
 \]
Speed of the sound \[\left( {{v_w}} \right)\] in water is given as \[1486\,{\text{m}}{{\text{s}}^{ - {\text{1}}}}\] .
The wavelength of the transmitted sound is related to the frequency by the following expression:
\[{\lambda _t} = \dfrac{{{v_w}}}{n}\] …… (2)
Where,
\[{\lambda _t}\] indicates the wavelength of the transmitted sound.
\[{v_w}\] indicates the velocity of sound in water.
\[n\] indicates the frequency of the emitted sound.

Substituting the values of frequency and velocity of the sound wave in equation (2), we get:
\[
 {\lambda _t} = \dfrac{{{v_w}}}{n} \\
 = \dfrac{{1486\,{\text{m}}{{\text{s}}^{ - 1}}}}{{{{10}^6}\,{\text{Hz}}}} \\
 = 1.486 \times {10^{ - 3}}\,{\text{m}} \\
\]

Hence, the wavelength of the reflected sound is found to be \[1.486 \times {10^{ - 3}}\,{\text{m}}\].

Note: To solve this problem, you should know that there are two different wavelengths associated with this phenomenon. It is important to note that for both the cases, frequency of the sound will remain the same, but there is a variation in the velocity of sound in both the media. Higher is the velocity, higher is the wavelength of the sound or vice-versa.