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# Two waves of wavelengths 2m and 2.02m respectively moving with the same velocity superpose produce 2 beats /sec. The velocity of the wave is:a) 400m/sb) 402m/sc) 404m/sd) 406m/s  Verified
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Hint: The two waves in the above question have a small difference in their respective wavelength. Hence they will superpose each other and travel as one. The velocity of a wave is given by $v=\lambda \gamma ...(1)$ where $\lambda$ is the wavelength of the wave and $\gamma$ is the frequency of the wave. Substituting the value of frequency in terms of wavelength of the wave and the and the unknown velocity of the wave, in the number of beats expression would give us the value of unknown velocity.

${{v}_{BEAT}}={{v}_{1}}-{{v}_{2}}....\text{(2),where }{{\text{v}}_{\text{1}}}\text{ and }{{\text{v}}_{\text{2}}}\text{ are the frequency of the two waves that are superposed}$.
${{v}_{1}}=\dfrac{\text{V}}{2m}\text{and }{{v}_{2}}=\dfrac{\text{V}}{2.02m}\text{ the velocity of both the waves i}\text{.e}\text{. V is the same since they are moving together}\text{.}$ 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}