Two sources of sound are emitting progressive | vedclass

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(C) The given equations are $y_1 = 4 \sin(710 \pi t)$ and $y_2 = 3 \sin(702 \pi t)$.Comparing these with $y = A \sin(2 \pi f t)$,we get:$2 \pi f_1 = 7

Vedclass · Accepted Answer

$4, 49:1$

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(C) The given equations are $y_1 = 4 \sin(710 \pi t)$ and $y_2 = 3 \sin(702 \pi t)$.Comparing these with $y = A \sin(2 \pi f t)$,we get:$2 \pi f_1 = 710 \pi \implies f_1 = 355 	ext{ Hz}$$2 \pi f_2 = 702 \pi \implies f_2 = 351 	ext{ Hz}$The number of beats per second is $n = |f_1 - f_2| = |355 - 351| = 4 	ext{ beats/s}$.The amplitudes are $A_1 = 4$ and $A_2 = 3$.The maximum intensity (waxing) is proportional to $(A_1 + A_2)^2 = (4 + 3)^2 = 7^2 = 49$.The minimum intensity (waning) is proportional to $(A_1 - A_2)^2 = (4 - 3)^2 = 1^2 = 1$.Therefore,the intensity ratio is $49:1$.The correct answer is $4, 49:1$.

Two sources of sound are emitting progressive waves $y_1 = 4 \sin(710 \pi t)$ and $y_2 = 3 \sin(702 \pi t)$. The sources are placed close to each other. The number of beats heard per second and the intensity ratio between waxing and waning are respectively:

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