Question: An oscillator vibrating at 1250 Hz produces a sound wave that travels through an ideal gas at 325 m/s when the gas temperature is 22.0° C. For a certain experiment, you need to have the same oscillator produce sound of wavelength 28.5 cm in this gas. What should the gas temperature be to achieve this wavelength?

The given data cen be listed as,

• The frequency of the vibrator is, \(f = 1250\;{\rm{Hz}}\).
• The velocity of the sound wave is, \(v = 325\;{\rm{m/s}}\) .
• The temperature that produces wavelength \({\lambda _1}\) is given as,\({T_1} = 22^\circ \;{\rm{C}} = \left( {22 + 273} \right)\;{\rm{K}} = 295\;{\rm{K}}\).
• The wavelength for temperature \({T_2}\) is given as,\({\lambda _2} = 28.5\;{\rm{cm}} = 28.5 \times {10^{ - 2}}\;{\rm{m}}\).

The expression for the wavelength of the sound wave traveling in a medium can be written as,

\(\lambda = \frac{1}{f} \times \sqrt {\frac{{\gamma RT}}{M}} \) (1)

Here\(\gamma \)is the adiabatic constant for the medium,\(R\)is the gas constant,\(T\)is the temperature,\(f\)is the frequency of the wave, and \(M\)is the molar mass of the medium.

The wavelength of these waves can be calculated using the expression,

\(\lambda = \frac{v}{f}\)

Here v is the velocity of the sound wave.

For temperature, \(295\;{\rm{K}}\)we can calculate the wavelength as,

\(\begin{array}{c}{\lambda _1} = \frac{{325\;{\rm{m/s}}}}{{1250\;{\rm{Hz}}}}\\ = 26 \times {10^{ - 2}}\;{\rm{m}}\end{array}\)

From equation 1, we can write the expression for \({\lambda _1}\) and \({T_1}\) as,

\({\lambda _1} = \frac{1}{f} \times \sqrt {\frac{{\gamma R{T_1}}}{M}} \) (2)

From equation 1, we can write the expression for \({\lambda _2}\) and \({T_2}\) as,

\({\lambda _2} = \frac{1}{f} \times \sqrt {\frac{{\gamma R{T_2}}}{M}} \) (3)

We need to find \({T_2}\), for that, let us divide equation 3 by equation 2 as,

\(\begin{array}{c}\frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{\frac{1}{f} \times \sqrt {\frac{{\gamma R{T_2}}}{M}} }}{{\frac{1}{f} \times \sqrt {\frac{{\gamma R{T_1}}}{M}} }}\\\frac{{{\lambda _2}}}{{{\lambda _1}}} = \sqrt {\frac{{{T_2}}}{{{T_1}}}} \end{array}\)

Substitute the values in the above expression, and we get,

\(\begin{array}{c}\frac{{28.5 \times {{10}^{ - 2}}\;{\rm{m}}}}{{26 \times {{10}^{ - 2}}\;{\rm{m}}}} = \sqrt {\frac{{{T_2}}}{{295\;{\rm{K}}}}} \\\sqrt {\frac{{{T_2}}}{{295\;{\rm{K}}}}} = 1.096\\\frac{{{T_2}}}{{295\;{\rm{K}}}} = 1.20\\{T_2} = 354.45\;{\rm{K}}\end{array}\)

Thus, the gas temperature should be \(354.45\;{\rm{K}}\) to achieve this wavelength \(28.5 \times {10^{ - 2}}\;{\rm{m}}\).