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 $t=\frac{1}{16 \, } \, s$ will be

• A. zero
• B. $\pi $
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (C)

The time interval between two consecutive beats
$T=\frac{1}{n_{1} - n_{2}}=\frac{1}{260 - 256}=\frac{1}{4}s$
So, $t=\frac{1}{16}=\frac{T}{4}s$
By using,
time difference $=\frac{T}{2 \pi }\times $ Phase difference
$\Rightarrow \, \, \frac{T}{4}=\frac{T}{2 \pi }\times \phi$
$\Rightarrow \, \, \phi=\frac{\pi }{2}$