Question

Velocity of sound in gas is $498m/s$ and in air $332m/s$. What is the ratio of wavelength of sound waves in gas to that in air?

Answer 1

We should solve this problem by using the relation of velocity of sound waves with frequency and wavelength of sound waves. We should ponder on the fact that velocity and wavelength of a given sound wave do change in different media but frequency remains the same as it does not depend on medium. Apply basic concepts of wavelength and velocity of sound waves.
Formula used:
We are using the following relation to solve this problem:-
$v=\lambda \nu$

Complete answer:
From the problem above, we get that the speed of sound in gas is, ${{v}_{1}}=498m/s$
And the speed of sound waves in air is, ${{v}_{2}}=332m/s$. We also know that the frequency, $\nu$ of sound waves does not depend on medium and therefore, remains the same in both gas and in air.
Now, using the relation,
$v=\lambda \nu$,
From the equation above we get the simplified relation,
$\lambda =\dfrac{v}{\nu }$
For gas we have,
${{\lambda }_{1}}=\dfrac{{{v}_{1}}}{\nu }$, where ${{\lambda }_{1}}$is wavelength of sound in gas.
Putting values, we get
${{\lambda }_{1}}=\dfrac{498}{\nu }$……………. (i)
For air we have,
${{\lambda }_{2}}=\dfrac{{{v}_{2}}}{\nu }$, where${{\lambda }_{2}}$is wavelength of sound in air.
Putting values, we get
${{\lambda }_{2}}=\dfrac{332}{\nu }$……………. (ii)
Dividing equation (i) by equation (ii), we get,
$\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{498}{332}$
$\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{3}{2}$
Therefore, the required ratio of wavelength of sound waves in gas and that in the air as $3:2$.

Additional Information:
Sound waves are mechanical waves as they require medium to travel. Speed of sound in different media is different. Sound travels faster in denser medium as compared to rarer medium at a given temperature. Speed of sound is temperature dependent also.

Note:
Proper concept of propagation of sound waves in different media is required in this problem. Always use the correct equations. Don’t be confused with the letters $v$ and $\nu$, as they look similar but represent velocity and frequency respectively.