Question

Two loudspeakers are placed side by side a distance d = 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .

Answer 1

Answer:

351 m/s

Explanation:

Given information

Distance between speakers, d= 4 m

Distance between listener and speaker, l= 5 m

Distance moved by the speaker, r= 60 cm = 0.6 m

Frequency, f= 700 Hz

$r_1=\sqrt {(d/2)^{2}+l^{2}}=\sqrt{(4/2)^{2}+5^{2}}=\sqrt{29}\approx 5.385 m$

$r_2=\sqrt {(r+d/2)^{2}+l^{2}}=\sqrt {(0.6+4/2)^{2}+5^{2}}\approx 5.636 m$

The path difference between two sound waves reaching the listener is half the integral of wavelength for destructive interference to happen hence

$r_2-r_1=\frac {\lambda}{2}$

$5.636-5.385=\frac {\lambda}{2}$

$\lambda= 0.502$

Speed of sound, $v=f\lambda= 700 Hz \times 0.502 m= 351 m/s$