Solved: PHYSICS Question for NEET

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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

$4$ beats per second with intensity ratio $25:16$ between waxing and waning

$8$ beats per second with intensity ratio $25:16$ between waxing and waning

$8$ beats per second with intensity ratio $81:1$ between waxing and waning

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

Step-by-Step Solution

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: $\omega_1 = 600\pi \implies 2\pi f_1 = 600\pi \implies f_1 = 300 \text{ Hz}$ For the second wave: $\omega_2 = 608\pi \implies 2\pi f_2 = 608\pi \implies f_2 = 304 \text{ Hz}$
Calculate Beat Frequency: The beat frequency is the difference between the two frequencies : $f_{\text{beat}} = |f_1 - f_2| = |300 - 304| = 4 \text{ beats/s}$
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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