Question

The speed of sound in a gas is in which two sound waves of wavelength $1.04m$ and $1.05m$ produces $15$ beats in $5$ sec is
A) $328m/s$
B) $330m/s$
C) $332m/s$
D) $320m/s$

Answer 1

To solve this question, we have to take into account the interference of sound waves. In the question, two sound waves are given along with their wavelengths. First, we have to find their respective frequencies. Now, after interference, the sound waves will produce a beat frequency whose value is given. We just have to use the formula of beat frequency to find the speed of sound in the gas. As the medium for the propagation of both the sound waves is the same, therefore, their speeds will also be the same.

Formula used:
$f = \dfrac{v}{\lambda }$
Where,
$f$ is the frequency of the sound wave.
$v$ is the speed of the sound wave.
$\lambda$ is the wavelength of the sound wave.
${f_b} = \left| {{f_2} - {f_1}} \right|$
Where,
${f_b}$ is the beat frequency.
${f_2}\,\& \,{f_1}$ are the frequencies of the sound waves before interference.

Complete answer:
Let the wavelengths of the two waves be ${\lambda _1}$ and ${\lambda _2}$ having frequencies ${f_1}$ and ${f_2}$ respectively.
In the question, the values of the wavelengths are given as:
${\lambda _1} = 1.04m$
${\lambda _2} = 1.05m$
So using the wave equation we get,
$f = \dfrac{v}{\lambda }......(1)$
Where,
$f$ is the frequency of the sound wave.
$v$ is the speed of the sound wave.
$\lambda$ is the wavelength of the sound wave.
We put the values of wavelength in equation (1) to get,
${f_1} = \dfrac{v}{{{\lambda _1}}}$
$\Rightarrow {f_1} = \dfrac{v}{{1.04}}......(2)$
Similarly, for the second wave, the frequency can be calculated as,
${f_2} = \dfrac{v}{{{\lambda _2}}}$
$\Rightarrow {f_2} = \dfrac{v}{{1.05}}......(3)$
Now, we have to find the beat frequency. It is given that the waves produce $15$ beats in $5$ sec. As the frequency is defined as the number of oscillations in one second. So to find the beat frequency, ${f_b}$ we will have to calculate the number of beats in a second. Therefore, we get
${f_b} = \dfrac{{15}}{5}$
${f_b} = 3Hz.....(4)$
But, mathematically the formula of beat frequency is:
${f_b} = \left| {{f_2} - {f_1}} \right|......(5)$
Where,
${f_b}$ is the beat frequency.
${f_2}\,\& \,{f_1}$ are the frequencies of the sound waves before interference.
Putting the values of the two frequencies along with the beat frequency from equations (2), (3), and (4) into equation (5) we get,
${f_b} = \left| {\dfrac{v}{{1.05}} - \dfrac{v}{{1.04}}} \right|$
$\Rightarrow 3 = \left| {\dfrac{v}{{1.05}} - \dfrac{v}{{1.04}}} \right|$
By taking the LCM and solving the LHS we get,
$\Rightarrow 3 = \left| {\dfrac{{1.04v - 1.05v}}{{1.04 \times 1.05}}} \right|$
$\Rightarrow 3 = \left| {\dfrac{{ - 0.01v}}{{1.092}}} \right|$
By taking the modulus and cross-multiplication, we will get,
$\Rightarrow 3 = \dfrac{{0.01v}}{{1.092}}$
$\Rightarrow v = \dfrac{{3 \times 1.092}}{{0.01}}$
$\Rightarrow v = 327.6m/s$
As the options given don’t match with the derived value, we have to approximate it to match. Hence, the speed of the sound in the gas is:
$v \approx 328m/s$
Hence the speed of sound in the gas is $328m/s$ and therefore option (A) is correct.

Note:
The sound waves propagate through the molecules present in the air which acts as a medium. If there is no medium, then the waves will not be able to travel. This is the reason why the astronauts cannot hear anything in space.