Saved Bookmarks
| 1. |
Two identical loudspeakers placed 3.00 m apart are driven by the same oscillator as shown in Fig. 7.12. A listener is originally at point O , located 8.00 m from the centre of the line connecting the two speakers. The listener then moves to point P, which is a perpendicular distance 0.350 m from O , and she experiences the first minimum sound intensity . What is the frequency of the oscillator ? |
|
Answer» Solution :In this example , signal representing the sound is electrically SPLIT and sent to two different loudspeakers . After leaving the speakers , the sound waves recombine at the positionof the listener. Despite the difference in how thesplitting occurs , the path differnce discussion related to Fig . 7.12 can be applied here. Because the sound waves from two separate sources combine , we apply the waves in interference analysis model. Figure 7.12 shows the physical arrangement of the speakers , along with two shaded right triangles that can be drawn on the basis of the lengths described in the problem . The first minimum occurs when the two waves reaching the listener at point `P are 180^(@)` out of phasw , in other words , when their path difference `Delta R equals lambda//2`. From the shaded triangles , find the path lengths from the speakers to the listener : `r_(1) = sqrt (( 8.00 m)^(2) + ( 1.15 m)^(2)) = 8.08 m` `r_(2)= sqrt(( 8.00 m)^(2) + ( 1.85 m)^(2)) = 8.21 m` Hence, the path difference is ` r_(2) - r_(1) = 0.132 m`. Because this path differnce must equal `lambda//2` for the first minimum , `lambda = 0.26 m`. To OBTAIN the oscillator FREQUENCY , use equation `v = lambda f`, where ` v` is the speed of sound in AIR , `343 m//s`. |
|