The frequencies of two sound sources are 256 | vedclass

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(C) The beat frequency is given by $f_b = |n_1 - n_2| = |260 - 256| = 4 	ext{ Hz}$.The time period of the beat is $T = \frac{1}{f_b} = \frac{1}{4}

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$\pi / 2$

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(C) The beat frequency is given by $f_b = |n_1 - n_2| = |260 - 256| = 4 	ext{ Hz}$.The time period of the beat is $T = \frac{1}{f_b} = \frac{1}{4} 	ext{ s}$.At $t = 0$,the intensity is maximum,which corresponds to a phase of $0$ or $2\pi$.The phase difference $\phi$ at time $t$ is related to the beat period $T$ by the formula $\phi = \frac{2\pi}{T} 	imes t$.Substituting the given values: $\phi = \frac{2\pi}{(1/4)} 	imes \frac{1}{16} = 8\pi 	imes \frac{1}{16} = \frac{\pi}{2}$ radians.

The frequencies of two sound sources are $256 \text{ Hz}$ and $260 \text{ Hz}$. At $t = 0$,the intensity of sound is maximum. Then the phase difference at the time $t = \frac{1}{16} \text{ s}$ will be

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