Question

The frequencies of two sound sources are 256 Hz and 260 Hz. At t = 0, the intensity of sound is maximum. Then the phase difference at the time t = 1/16 sec will be

• A. Zero
• B. π
• C. π/2
• D. π/4

Answer 1

Answer: (C)

π/2

Answer 2

Time interval between two consecutive beats

$T = \frac{1}{n_{1} - n_{2}} = \frac{1}{260 - 256} = \frac{1}{4}\sec$ so, $t = \frac{1}{16} = \frac{T}{4}\sec$

By using time difference =$\frac{T}{2\pi} \times$Phase difference