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Question: The wavelengths of two waves are 90 cm and 91 cm. If the room temperature is \({45^\circ }C\) then t...

The wavelengths of two waves are 90 cm and 91 cm. If the room temperature is 45C{45^\circ }C then the number of beats produced per second by these waves will be (velocity of sound at 0C{0^\circ }C= 333 m/s).
(A) 4.4
(B) 10
(C) 8
(D) 12

Explanation

Solution

Hint To answer this question we should at first write down the formula which relates the velocities and the temperature. Then we have to put the values that are mentioned in the question. Finally, to find the frequencies we have to consider the values of the wavelengths that are mentioned in the question. This will give us the required answer.

Complete step by step answer
We know that the velocity of sound in air is v. This can be written as vαTv\alpha \sqrt T.
Therefore, we can write that:
vTv0=T2T1\Rightarrow \dfrac{{{v_T}}}{{{v_0}}} = \sqrt {\dfrac{{{T_2}}}{{{T_1}}}}
We know that it is given:
v0=333m/s,T1=273K,T2=273+45=318K{v_0} = 333m/s,{T_1} = 273K,{T_2} = 273 + 45 = 318K
We can now write that:
vT333=318273\dfrac{{{v_T}}}{{333}} = \sqrt {\dfrac{{318}}{{273}}}
vT=359.7m/s=35927cm/s\Rightarrow {v_T} = 359.7m/s = 35927cm/s
It is also given that:
λ1=90cm,λ2=91cm{\lambda _1} = 90cm,{\lambda _2} = 91cm
We know that beat frequency is b. The formula of b is given as:
b=v1v2=vT(1λ11λ2)b = {v_1} - {v_2} = {v_T}\left( {\dfrac{1}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right)
Therefore, now we have to put the values in the equation:
b=(35927)(190191)=4.4Hzb = (35927)\left( {\dfrac{1}{{90}} - \dfrac{1}{{91}}} \right) = 4.4Hz
So we can say that the wavelengths of two waves are 90 cm and 91 cm. If the room temperature is 45C{45^\circ }C then the number of beats produced per second by these waves will be 4.4 Hz.

Hence the correct answer is Option A.

Note We have come across the term beat frequency in the answer. To avoid getting any confusion we must define the term beat frequency. By beat frequency is defined as the rate in which the volume is heard to be oscillating from a high volume to a low volume.
The main reason behind the cause of the beat frequency is the interference that occurs in the wave due to the similar frequencies.