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 ) s will be

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 ) s will be (A) zero (B) π  (C) (π /2) (D) (π /4)

Tardigrade Admin · Accepted Answer

The time interval between two consecutive beats T=(1/n1 - n2)=(1/260 - 256)=(1/4)s So, t=(1/16)=(T/4)s By using, time difference =(T/2 π )× Phase difference ⇒ (T/4)=(T/2 π )× φ ⇒ φ=(π /2)

Q. 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

zero

$π$

$\frac{π}{2}$

$\frac{π}{4}$

Q. The frequencies of two sound sources are 256Hz and 260Hz . At t=0 , the intensity of sound is maximum. Then the phase difference at the time t=161​s will be

zero

π

2π​

4π​

The time interval between two consecutive beats T=n1​−n2​1​=260−2561​=41​s So, t=161​=4T​s By using, time difference =2πT​× Phase difference ⇒4T​=2πT​×ϕ ⇒ϕ=2π​

Q. 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

$π$

$\frac{π}{2}$

$\frac{π}{4}$