Physics - Sound Wave Question with Solution | TestHub

Given equations of progressive waves:

$y_1=4\sin (600\pi t)$

$y_2=5\sin (608\pi t)$

Comparing with the standard equation $y=A\sin (\omega t)$:

$A_1=4$, ${\omega }_1=600\pi \Rightarrow f_1=\frac{600\pi }{2\pi }=300$ Hz

$A_2=5$, ${\omega }_2=608\pi \Rightarrow f_2=\frac{608\pi }{2\pi }=304$ Hz

The beat frequency is given by:

$n=|f_2-f_1|=304-300=4$ beats per second

The ratio of maximum to minimum intensity (waxing to waning) is:

$\frac{I_{max}}{I_{min}}={\left(\frac{A_1+A_2}{A_1-A_2}\right)}^2$

$\frac{I_{max}}{I_{min}}={\left(\frac{4+5}{4-5}\right)}^2={\left(\frac{9}{-1}\right)}^2=\frac{81}{1}$

Thus, the observer hears $4$ beats per second with an intensity ratio of $81:1$.

Answer: $4$ beats per second with an intensity ratio of $81:1$ between waxing and waning