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a) 400m/s

b) 402m/s

c) 404m/s

d) 406m/s

Answer

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Let us first define what is beat frequency or beats.

When two waves having small differences in their frequency superpose each other there is a periodic variation in the intensity of the sound i.e. the amplitude of the sound rises and falls. The number of such variations heard per second is called the beat frequency.

The beat frequency is given by,

${{v}_{BEAT}}={{v}_{1}}-{{v}_{2}}....\text{(2),where }{{\text{v}}_{\text{1}}}\text{ and }{{\text{v}}_{\text{2}}}\text{ are the frequency of the two waves that are superposed}$.

In the above question the wavelengths of the two superposing waves are given to us. Let us write the wavelength in terms of their frequency using equation 1.

${{v}_{1}}=\dfrac{\text{V}}{2m}\text{and }{{v}_{2}}=\dfrac{\text{V}}{2.02m}\text{ the velocity of both the waves i}\text{.e}\text{. V is the same since they are moving together}\text{.}$ Now let us substitute the frequency of both the waves in equation 2 to determine the velocity of the wave.

$\begin{align}

& {{v}_{BEAT}}={{v}_{1}}-{{v}_{2}} \\

& {{v}_{BEAT}}=\dfrac{\text{V}}{2m}-\dfrac{\text{V}}{2.02m} \\

\end{align}$ we know that the beat frequency is 2beats/sec. Hence

$2=\text{V}\left( \dfrac{1}{2m}-\dfrac{1}{2.02m} \right)$

$\begin{align}

& 2=\text{V}\left( 0.5-0.495049 \right) \\

& \text{V}=\dfrac{2}{4.95\times {{10}^{-3}}}=0.404\times {{10}^{3}}=404\text{m/s} \\

\end{align}$