Two vibrating tuning forks produce progressiv | vedclass

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(B) The given equations for the waves are $y_1 = 4 \sin(500 \pi t)$ and $y_2 = 2 \sin(506 \pi t)$.Comparing these with the standard form $y = A \sin(2

Vedclass · Accepted Answer

$3 	ext{ beats/s}$ with intensity ratio between maxima and minima equal to $9$.

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(B) The given equations for the waves are $y_1 = 4 \sin(500 \pi t)$ and $y_2 = 2 \sin(506 \pi t)$.Comparing these with the standard form $y = A \sin(2 \pi n t)$,we get:For the first fork: $2 \pi n_1 = 500 \pi \implies n_1 = 250 	ext{ Hz}$.For the second fork: $2 \pi n_2 = 506 \pi \implies n_2 = 253 	ext{ Hz}$.The number of beats per second is given by $|n_2 - n_1| = |253 - 250| = 3 	ext{ beats/s}$.The amplitudes are $A_1 = 4$ and $A_2 = 2$. Since intensity $I \propto A^2$,we have $I_1 \propto 16$ and $I_2 \propto 4$.The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} ight)^2$.Substituting the values: $\frac{I_{\max}}{I_{\min}} = \left( \frac{4 + 2}{4 - 2} ight)^2 = \left( \frac{6}{2} ight)^2 = 3^2 = 9$.Thus,the person will hear $3 	ext{ beats/s}$ with an intensity ratio of $9$.

Two vibrating tuning forks produce progressive waves given by $y_1 = 4 \sin(500 \pi t)$ and $y_2 = 2 \sin(506 \pi t)$. These tuning forks are held near the ear of a person. The person will hear

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