Question

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/s
b) 402m/s
c) 404m/s
d) 406m/s

Answer 1

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.

Complete step by step answer:
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
$2=\text{V}\left( \dfrac{1}{2m}-\dfrac{1}{2.02m} \right)$
$\begin{align}
  & 2=\text{V}\left( 0.5-0.495049 \right) \\
 & \text{V}=\dfrac{2}{4.95\times {{10}^{-3}}}=0.404\times {{10}^{3}}=404\text{m/s} \\
\end{align}$

So, the correct answer is “Option C”.

Note: The two waves having minute difference in their frequencies only superpose each other. The difference should not be more than of order 10.This is because if the number of beats produced is more than 10 the beats would not be heard due to human persistence of hearing.