Two identical coherent sound sources R and S | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a minimum intensity at point $P$ (directly in front of source $R$),the path difference between the sound waves reaching $P$ from $S$ and $R$ m

Vedclass · Accepted Answer

$495$

Vedclass · Answer

(A) For a minimum intensity at point $P$ (directly in front of source $R$),the path difference between the sound waves reaching $P$ from $S$ and $R$ must be an odd multiple of half the wavelength.The path difference is given by $\Delta x = SP - RP$.Given the geometry,$RP = 12 \,m$ and $RS = 5 \,m$. Thus,$SP = \sqrt{RP^2 + RS^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \,m$.The path difference is $\Delta x = 13 \,m - 12 \,m = 1 \,m$.For destructive interference (minima),$\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, \dots$.Substituting $\lambda = \frac{v}{f}$,we get $1 = (2n + 1) \frac{v}{2f}$,which implies $f = \frac{(2n + 1)v}{2}$.Given $v = 330 \,m/s$,we have $f = \frac{(2n + 1) 	imes 330}{2} = (2n + 1) 	imes 165$.For $n = 0$,$f = 165 \,Hz$.For $n = 1$,$f = 3 	imes 165 = 495 \,Hz$.For $n = 2$,$f = 5 	imes 165 = 825 \,Hz$.Comparing with the given options,$495 \,Hz$ is a possible value.

Similar Questions

Two waves are superimposed whose ratio of intensities is $9: 1$. The ratio of maximum and minimum intensity is

Two waves have intensities $x$ and $y$. If the time difference between them is $3T/2$,what is the resultant intensity?

Give noticeable points about the amplitude of resultant waves of two harmonic progressive waves on a stretched string.

Two waves of the same frequency and same amplitude $a$ are superimposed. If the resultant amplitude is also $a$,what is the phase difference between the two waves?

State the principle of superposition and explain it.

Vedclass Test Series

Exam Paper Generator

Online Exam Module