Two loudspeakers are placed at either end of a gymnasium, both pointing toward the center of the gym and equidistant from it. The speakers emit 296 Hz sound that is in phase. An observer at the center of the gym experiences constructive interference. How far toward either speaker must the observer walk to first experience destructive interference?

The observer must walk 0.2897m towards either speaker.

Explanation:

Let us make distance between the two speakers to be =D

frequency emitted by the two speakers in phase=f

v= speed of sound, 343 m/s

λ=wavelength

solving for λ, wavelength, using the equation λ=v/f. 343/296

= 1.15878m

Also destructive interference is given as

⎮d1-d2⎮=λ/2

Because the observer should be able to walk far enough to make the distance to one speaker differ from the distance to the other speaker by ½ wavelength.

This will give us

d1-d2= 1.15878 /2

d1-d2=0.5794m

For the distance moved on either side, we have D/2

Therefore 0.5794/2=distance to move on either side. Giving us

0.2897m

In conclusion, This means that the observer has to walk so as to be 0.5794m closer to one speaker from another; this means walking 0.2897m towards either speaker.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

current flows in a wire