Question

The speed of sound in a gas, in which two waves of wavelength $1.0m$ and $1.02m$ produce $6$ beats per second, is approximately:
(A) $350m/s$
(B) $300m/s$
(C) $380m/s$
(D) $410m/s$

Answer 1

Hint
To find the speed of sound we can use the wave equation, where the given beat frequency is the difference between the frequencies of the two waves, and the wavelengths of the two waves are given in the question. The speed of sound for both the waves will be the same.
Formula Used:
$n = \dfrac{v}{\lambda }$
where $n$ is the frequency of the sound wave
$v$ is the speed of the sound wave in that medium
$\lambda$ is the wavelength of the sound waves
${n_1} - {n_2}$ is beat frequency
where ${n_1}$ is the frequency of the first wave
${n_2}$ is frequency of the second wave

Complete step by step answer
Here in the problem, we can see that the wavelengths of 2 waves are given as $1.0m$ and $1.02m$ .
$\therefore {\lambda _1} = 1.0m{\text{, }}{\lambda _2} = 1.02m$
We consider ${n_1}$ the frequency of the first wave and ${n_2}$ the frequency of the second wave.
So we can write two wave equations for the two waves,
${n_1} = \dfrac{v}{{{\lambda _1}}}$ and ${n_2} = \dfrac{v}{{{\lambda _2}}}$
Here the medium is the same in both the cases so the speed of the sound wave is $v$ in both the cases.
Now we know that the difference between the frequencies of the two waves gives the wave frequency.
So, ${n_1} - {n_2} = 6$
Now we can put the values of ${n_1}$ and ${n_2}$ in the above equation from the wave equation,
$\dfrac{v}{{{\lambda _1}}} - \dfrac{v}{{{\lambda _2}}} = 6$
Substituting the values of ${\lambda _1}$ and ${\lambda _2}$ and then taking the value of $v$ as common from both the terms is the L.H.S we get,
$v\left( {\dfrac{1}{{1.0}} - \dfrac{1}{{1.02}}} \right) = 6$
Now we are taking doing the fraction subtraction, and get
$v\left( {\dfrac{{1.02 - 1}}{{1.02}}} \right) = 6$
By taking the fraction over to the R.H.S we get,
$\Rightarrow v = 6 \times \dfrac{{1.02}}{{0.02}}$
By calculating this we get,
$\Rightarrow v = 306m/s$
Now we have to take the approximate value of $v$ as in the question. So, therefore we get
$v \simeq 300m/s$ .
Hence the correct answer is $300m/s$ , and the correct option is (B).

Additional Information
Though the speed of a wave depends on the medium, as the temperature of the medium increases, the corresponding speed of the wave also increases. The frequency of a wave changes with the energy of the wave. The higher the energy the higher is the frequency of that wave. Since in a given medium the speed is constant, from the wave equation we can say that a high-frequency wave has a shorter wavelength and vice-a-versa.

Note
In the solution we have taken the value of $v$ as the same in both the wave equations. This is because the speed of sound solely depends on the medium and the temperature of the medium. Since the medium is the same in both cases, so the speed remains the same. In the given problem if we had taken ${\lambda _1} = 1.02m{\text{ and }}{\lambda _2} = 1.0m$ then by doing the same calculation we would have gotten the value of $v$ as $v = - 306m/s$ . As the value of speed cannot be negative so in that case, we would have taken $\left| v \right| = 306m/s$ .