MCQ
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 $t = \frac{1}{16}\, sec$ will be
- AZero
- B$\pi$
- ✓$\pi /2$
- D$\pi/4$
✓
Answer
Correct option: C.
$\pi /2$
c
(c) 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
$ \Rightarrow $$\frac{T}{4} = \frac{T}{{2\pi }} \times \phi \Rightarrow \phi = \frac{\pi }{2}$
(c) Time interval between two consecutive beats
$T = \frac{1}{{{n_1} - {n_2}}} = \frac{1}{{260 - 256}} = \frac{1}{4}\sec $ so, $t = \frac{1}{{16}} = \frac{T}{4}\,\,\sec $
By using time difference =$\frac{T}{{2\pi }} \times $Phase difference
$ \Rightarrow $$\frac{T}{4} = \frac{T}{{2\pi }} \times \phi \Rightarrow \phi = \frac{\pi }{2}$
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