Two loudspeakers placed 8.0 m apart are driven in phase by an audio oscillator whose frequency range is 2.2 kHz to 2.9 kHz. A point P is located 5.4 m from one loudspeaker and 4.6 m from the other. The speed of sound is 344 m/s.

The answer to the question is 2.2khz

Explanation:

Let z₁ = 5.4m

Let z₂ = 4.6m

The path difference Δz = z₁-z₂ = 5.4 - 4.6 = 0.8m

For the interference= Δz λ, 2λ, 3λ

The wavelength λ = 0.8m

The speed of sound v = 344m/s

The frequency f = v/λ = 344/0.8 = 430hz

Now,

f₁ =f, f₂= 2f, f₃ = 3f, f₄= 4f, f₅ =5f which is,

f₁ =f = 430Hz, f₂=2f =860Hz, f₃ =3f =1290Hz f₄ =4f =1720Hz and f₅=5f =2150Hz

f5 = 2120Hz = 2.200Hz

we will convert to two significant figures =2.2kHz

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.