PROBLEM 14.2: "If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. The object is displaced a distance 0.120 m from its equilibrium position and released with zero initial speed. Then after a time 0.800 s, its displacement is found to be a distance 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. a) Find the amplitude. b) Find the period. c) Find the frequency."

a) , b) , c)

Explanation:

The known conditions of the motion are:

a) The amplitude is the half of the difference between both positions:

b) The period of oscillation is equal to the double of the time between both positions:

c) The frequency of the oscillation is:

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.