Question

The velocity of sound in a gas is 4 times that in air at the same temperature. When a tuning fork is sounded in a wave of frequency 480 Hz and wavelength ${\lambda _1}$ is produced. The same fork is sounded in the gas and ${\lambda _2}$ is the wavelength of the wave, then ${\lambda _2}/{\lambda _1}$ is:
A. 1
B. 2
C. 4
D. 1/2

Answer 1

The sound is a longitudinal wave and therefore, it has both frequency and wavelength. Recall the relationship between velocity, frequency and wavelength and express the velocity of sound in the air and in the gas. Substituting ${v_g} = 4{v_a}$ in the equation, you will get the ratio ${\lambda _2}/{\lambda _1}$.

Formula used:
$v = f\lambda$
Here, v is the velocity, f is the frequency and $\lambda$ is the wavelength of the sound wave.

Complete step by step answer:
We have given that the velocity of sound in a gas is 4 times that in air. Therefore, ${v_g} = 4{v_a}$, where, ${v_g}$ is the velocity of the sound in the gas and ${v_a}$ is the velocity of sound in the air. We have the relation between velocity, frequency and wavelength,
$v = f\lambda$
Here, v is the velocity, f is the frequency and $\lambda$ is the wavelength of the sound wave.

Let us express the velocity of the sound wave in the air as follows,
${v_a} = f{\lambda _1}$ …… (1)
Here, ${\lambda _1}$ is the wavelength of the sound wave in air.
Let us express the velocity of the sound wave in the gas as follows,
${v_g} = f{\lambda _2}$ …… (2)
Here, ${\lambda _2}$ is the wavelength of the sound wave in the gas.
The frequency of the wave is the same since the source of the sound that is tuning the fork is the same in both the cases.

Dividing equation (2) by equation (1), we get,
$\dfrac{{{v_g}}}{{{v_a}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}$ …… (3)
Substituting ${v_g} = 4{v_a}$ in the above equation, we get,
$\dfrac{{4{v_a}}}{{{v_a}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}$
$\therefore \dfrac{{{\lambda _2}}}{{{\lambda _1}}} = 4$
Therefore, the ratio $\dfrac{{{\lambda _2}}}{{{\lambda _1}}}$ is 4.

So, the correct answer is option C.

Note: Students must know that the frequency of the sound wave remains the same even when the sound travels through a different medium. The frequency of the sound depends on the vibrations of the source which is creating it. The speed of light and speed of sound both changes when they travel through different mediums.