Question

(III) A source emits sound of wavelengths 2.54 m and 2.72 m in air. (a) How many beats per second will be heard? (Assume\(T = 20^\circ {\rm{C}}\)) (b) How far apart in space are the regions of maximum intensity?

Short answer

The beats per second is (a) \(8.9\;{\rm{beats/s}}\) and the distance is (b) \({\lambda _{\rm{B}}} = 38.5\;{\rm{m}}\).

Step-by-step solution

Given Data

The wavelength of sound is \(\lambda = 2.54\;{\rm{m}}\).

The other wavelength of sound is \(\lambda ' = 2.72\;{\rm{m}}\).

The temperature is \(T = 20^\circ {\rm{C}}\).

Understanding the beat frequency

In this problem, the beat frequency can be estimated by subtracting the frequency of each sound in air emitting from the source.

Evaluating the beat frequency

The relation to the calculate beat frequency is given by,

\(\begin{aligned}{l}{f_{\rm{B}}} &= \left| {{f_1} - {f_2}} \right|\\{f_{\rm{B}}} &= \left| {\frac{v}{\lambda } - \frac{v}{{\lambda '}}} \right|\end{aligned}\)

Here, vis the speed of the sound.

On plugging the values in the above relation.

\(\begin{aligned}{l}{f_{\rm{B}}} &= \left| {\frac{{343\;{\rm{m/s}}}}{{2.54\,{\rm{m}}}}\frac{{343\;{\rm{m/s}}}}{{2.72\;{\rm{m}}}}} \right|\\{f_{\rm{B}}} &= 8.9\;{\rm{beats/s}}\end{aligned}\)

Thus, \({f_{\rm{B}}} = 8.9\;{\rm{beats/s}}\) is the required beats per second.

Evaluating the distance between the regions of maximum intensity

The relation of wavelength is given by,

\({\lambda _{\rm{B}}} = \frac{v}{{{f_{\rm{B}}}}}\)

On plugging the values in the above relation.

\(\begin{aligned}{l}{\lambda _{\rm{B}}} &= \left( {\frac{{343\;{\rm{m/s}}}}{{8.9\;{\rm{beats/s}}}}} \right)\\{\lambda _{\rm{B}}} &= 38.5\;{\rm{m}}\end{aligned}\)

Thus, \({\lambda _{\rm{B}}} = 38.5\;{\rm{m}}\) is the required distance.