Question

The velocity of sound is ${V_S}$ in air. If the density of air is increased to $4$ times, the new velocity of sound is:
A) $\dfrac{{{V_S}}}{2}$
B) $\dfrac{{{V_S}}}{{12}}$
C) $12{V_S}$
D) $\dfrac{3}{2}{V_S}^2$

Answer 1

Hint: Here we have to apply the concepts of speed of travelling waves.
The speed of an object alludes to how quick an object is moving and is normally communicated as the distance travelled per time of travel. On account of a wave, the speed is the travelled by a given point on the wave, (for example, a peak) in a given time period.

Complete step-by-step answer:
As a wave is watched going through a medium, a peak is seen moving along from particle to particle. This sort of wave design that is seen going through a medium is now and then known as a travelling wave. In a longitudinal wave, the constituents of the medium moves to and fro toward propagation of the wave. We realize that the sound waves travel as compression and rarefactions of the little volume component of the medium.
Speed of longitudinal waves in a medium is given by:
$v = \sqrt {\dfrac{B}{\rho }} $
Where $B$ is the bulk modulus and $\rho $ is the density of the medium.
It can also be shown that the speed of a longitudinal wave in the bar is given by:
$v = \sqrt {\dfrac{Y}{\rho }} $
After applying isothermal change we get:
$\Rightarrow$$v = \sqrt {\dfrac{P}{\rho }} $
According to question,
${v_S} = \sqrt {\dfrac{P}{{{\rho _1}}}} $
$\Rightarrow$${V_S}' = \sqrt {\dfrac{P}{{{\rho _2}}}} $
$\Rightarrow$${\rho _2} = 4 \times {\rho _1}$
$ \dfrac{{{V_S}'}}{{{V_S}}} = \sqrt {\dfrac{{{\rho _1}}}{{{\rho _2}}}} \\
   = \sqrt {\dfrac{{{\rho _1}}}{{4 \times {\rho _2}}}} \\
   = \dfrac{1}{2} \\
$
$\Rightarrow$${V_S}' = \dfrac{{{V_S}}}{2}$

Hence, option A is correct.

Note: Here we have to pay attention as to how many times the density of air is increased. Also we have to remember the formula for the travelling wave correctly. Travelling waves are seen when a wave isn't restricted to a given space along the medium.