Two sources of sound placed close to each other, are emitting progressive waves given by $y_1 = 4 \sin(600\pi

Question

Two sources of sound placed close to each other, are emitting progressive waves given by $y_1 = 4 \sin(600\pi t)$ and $y_2 = 5 \sin(608\pi t)$. An observer located near these two sources of sound will hear

Accepted Answer

1. **Determine Frequencies:** Compare the given wave equations $y_1 = 4 \sin(600\pi t)$ and $y_2 = 5 \sin(608\pi t)$ with the standard wave equation $y = A \sin(\omega t)$, where $\omega = 2\pi f$.
   For the first wave: $\omega_1 = 600\pi \implies 2\pi f_1 = 600\pi \implies f_1 = 300 	ext{ Hz}$
   For the second wave: $\omega_2 = 608\pi \implies 2\pi f_2 = 608\pi \implies f_2 = 304 	ext{ Hz}$
2. **Calculate Beat Frequency:** The beat frequency is the difference between the two frequencies :
   $$f_{	ext{beat}} = |f_1 - f_2| = |300 - 304| = 4 	ext{ beats/s}$$
3. **Determine Amplitudes and Intensity Ratio:** The amplitudes of the waves are $A_1 = 4$ and $A_2 = 5$. The intensity of a wave is directly proportional to the square of its amplitude ($I \propto A^2$).
   - For waxing (constructive interference/maximum intensity): $I_{\max} \propto (A_1 + A_2)^2 = (4 + 5)^2 = 9^2 = 81$ .
   - For waning (destructive interference/minimum intensity): $I_{\min} \propto (A_1 - A_2)^2 = (4 - 5)^2 = (-1)^2 = 1$ .
   - The intensity ratio between waxing and waning is therefore $I_{\max} : I_{\min} = 81:1$.

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