Two identical loudspeakers are placed on a wall 3.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (Your answer should be between 0 and 2?.) (b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

a)  ΔФ = 5.54 rad , b)   f = 140 Hz  

Explanation:

a) This is a sound interference exercise, which is described by

         Δr /λ = ΔФ / 2π

         ΔФ = Δr 2π /λ

Let's find the path difference

       Δr = r₂ -r₁

       r₁ = 4m

       r₂ = √ (x² + y²) = √(3² + 4²) = 5 m

       Δr = 1 m

To find the wavelength we use the relation of the speed of sound

         v = λ f

         λ = v / f

         λ = 340/300

         λ = 1,133 m

We substitute

        ΔФ = 2π 1 /1.133

         ΔФ = 5.54 rad

b) to have a minimum intensity the phase difference must be π radians

         λ = Δr 2π /Ф

         λ = 1 2π /π

         λ = 2m

We look for the frequency

        v = λ f

        f = v /λ

        f = 340/2

        f = 140 Hz  

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.