Question

The wavelengths of two waves are 50 and 51 cm respectively. If the temperature of the room is \[{20^0}C\], then what will be the number of beats produced per second by these waves, when the speed of sound at \[{0^0}C\] is 332 m/s?
A. 14
B. 10
C. 24
D. none of these

Answer 1

Hint: The number of beats produced per second by the waves or beat frequency is defined as the difference in frequency of two waves. The frequency can be calculated by using the formula for speed of sound which is directly proportional to the square root of the temperature.

Formula used
Speed of propagation of sound waves related to the absolute temperature is given as,
\[v \propto \sqrt T \]
Where, v is speed of sound and T is absolute temperature measured in Kelvin scale.
Speed of sound is given as,
\[v = f\lambda \]
Where f is the frequency and \[\lambda \] is the wavelength.

Complete step by step solution:
Given wavelengths of two waves are 50 cm (or 50/100 m) and 51 cm (52/100 m).
At \[T = {0^0}C{\rm{ \,or\, 273K}}\], the speed of sound v = 332 m/s.
Let us assume that at the given temperature of the room \[{T^1} = {20^0}C\], the speed of sound is \[{v^1}\]. As we know that the speed of propagation of sound waves is related to the absolute temperature of the atmosphere is given by the relation,
\[v \propto \sqrt T \\ \]
Now we can write,
\[\dfrac{{{v^1}}}{v} = \sqrt {\dfrac{{{T^1}}}{T}} \\ \]
\[\Rightarrow {v^1} = 332\sqrt {\dfrac{{293}}{{273}}} \\ \]
\[\Rightarrow {v^1} = 344{\rm{ m/s}} \\ \]
As we know that speed of sound is given as
\[v = f\lambda \] or \[f = \dfrac{v}{\lambda } \\ \]
Using the values, we get
\[{f_1} = \dfrac{{{v^1}}}{{{\lambda ^1}}} = \dfrac{{344 \times 100}}{{50}} = 688{\rm{ Hz}} \\ \]
\[\Rightarrow {f_2} = \dfrac{v}{\lambda } = \dfrac{{344 \times 100}}{{51}} = 674{\rm{ Hz}}\]
Thus, the number of beats produced per second by the waves (or beat frequency) = \[{f_1} - {f_2}\] = 688-674 = 14 Hz

Hence option A is the correct answer.

Note: Sound waves are the longitudinal waves. Speed of a longitudinal wave when propagating in air is directly proportional to the temperature of air at the time of wave propagation.