Two sources of sound placed closed to each other, are emitting progressive waves given by
$y_{1} = 4 sin (600 π t)$ and $y_{2} = 5 sin (608 π t)$ .
An observer located near these two sources of sound will hear

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TestHub · Accepted Answer

The given equations of the progressive waves are:
$y_{1} = 4 sin (600 π t)$
$y_{2} = 5 sin (608 π t)$

Comparing these with the standard equation $y = A sin (ω t)$ , we get the angular frequencies:
$ω_{1} = 600 π \Rightarrow 2 π f_{1} = 600 π \Rightarrow f_{1} = 300$ Hz
$ω_{2} = 608 π \Rightarrow 2 π f_{2} = 608 π \Rightarrow f_{2} = 304$ Hz

The beat frequency is the difference between the frequencies of the two waves:
$n = ∣ f_{1} - f_{2} ∣ = ∣300 - 304∣ = 4$ beats/s

The amplitudes of the two waves are $A_{1} = 4$ and $A_{2} = 5$ .
The maximum amplitude during waxing is $A_{ma x} = A_{1} + A_{2} = 4 + 5 = 9$ .
The minimum amplitude during waning is $A_{min} = ∣ A_{1} - A_{2} ∣ = ∣4 - 5∣ = 1$ .

Since intensity is proportional to the square of the amplitude ( $I \propto A^{2}$ ), the intensity ratio between waxing and waning is:
$\frac{I _{ma x}}{I _{min}} = (\frac{A _{ma x}}{A _{min}})^{2} = (\frac{9}{1})^{2} = \frac{81}{1}$

Thus, the observer will hear $4$ beats/s with an intensity ratio of $81 : 1$ .

Answer: $4$ beats/s with intensity ratio $81 : 1$ between waxing and waning.

Physics - Sound Wave Question with Solution | TestHub

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