Question

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounded alone. The speakers are driven in-phase (at the speakers) with various frequencies of sound. The distance between the two speakers is 3m and the distance between the speaker S1 and the oberver is 4m..
Find the lowest frequencies, f1 for which the observer at O will hear an intensity of 16 W/m² when both speakers are on.

Answer 1

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$=3m

$S_1 O$=4m

By Pythagoras theorem

                                 $S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

• I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

• I1 is the intensity of one speaker on which is 6  $W/m^2$

• δ is the Path difference which is given as

                                           $\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

• λ is wavelength which is given as

                                             $\lambda=\frac{v}{f}$

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  $16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$, k=1

                                  ${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz