Question

Two loudspeakers $S_{1}and{S}_2$ each emit sounds of frequency 220 Hz uniformly in all directions. $S_1$ has an acoustic output of $1.2\times {10}^{-3}$ W and $S_2$ has $1.8\times {10}^{-3}$ W. $S_{1}and{S}_2$ vibrate in phase. Consider a point P such that $S_{1}P=0.75m$ and $S_{2}P=3m$. How are the phases arriving at P related? What is the intensity at P when both $S_{1}and{S}_2$ are on ? Speed of sound in air is 330 m/s.

Answer 1

Given - $P=1.2\times {10}^{-3}W,P^{\prime }=1.8\times {10}^{-3}W,f=220Hz,v=330m/s$

wavelength of the first speaker ,

$\lambda =v/f=330/220=1.5m$ ,

so wavelength of the second speaker ${\lambda }^{\prime }=1.5m$, (given that frequencies of both speakers are same) ,

for first speaker , $S_{1}P=0.75m=x$(path ),

therefore , phase $\varphi =\left(2\pi /\lambda \right)x=\left(2\pi /1.5\right)0.75=\pi$ ,

for second speaker , $S_{2}P=3m=x^{\prime }$(path ),

therefore , phase ${\varphi }^{\prime }=\left(2\pi /\lambda \right)x^{\prime }=\left(2\pi /1.5\right)3=4\pi$ ,

now path difference at P, $\Delta x=3-0.75=2.25m$ ,

hence , phase difference $\Delta {\varphi }^{\prime \prime }=\left(2\pi /1.5\right)2.5=3\pi$ ,

intensity of the first wave , $I=P/4\pi r^2=1.2\times {10}^{-3}/(4\times 3.14\times {0.75}^2)=17\times {10}^{-5}W/m^2$ ,

intensity of the second wave , $I^{\prime }=P^{\prime }/4\pi r^{\prime 2}=1.8\times {10}^{-3}/(4\times 3.14\times 3^2)=1.5\times {10}^{-5}W/m^2$ ,

therefore resultant intensity at point P ,

$I^{\prime \prime }=I+I^{\prime }+2\sqrt{I{I}^{\prime }}\cos \left({\varphi }^{\prime \prime }\right)=17\times {10}^{-5}+1.5\times {10}^{-5}+2\sqrt{17\times {10}^{-5}\times 1.5\times {10}^{-5}}\cos 3\pi$ ,

or $I^{\prime \prime }=18.5\times {10}^{-5}-10\times {10}^{-5}=8.5\times {10}^{-5}W/m^2$