What are (a) the amplitude and (b) the phase angle (relative to wave 1) of the resultant wave? (c) If a third wave of amplitude 4.64 mm is also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?

a) Ym = 3.29 mm  

b) ∅_R=85.5°=1.5 rad

c) If a third wave is also to be sent along the string in the same direction as

  the first two waves,then its phase angle must be in same as the resultant

  of the first 2 to maximise the amplitude.

Explanation:

Givens:

Ym_1= 4.6 mm

Ym_2 = 5.6 min,  

∅_1 = 0

∅_2 = 0.8  π

Remark:

remember the cosine ans sin rule we will use them as follows in cases of 2 waves only  

Ym^2=Ym_1^2+Ym_2^2-2*Ym_1*Ym_2*cos(π-∅_2)

sin(π-∅)/Ym=sin(∅_R)/Ym_2

So by substitution in both rules we got  

Part(a)

Ym^2= 4.62^2 + 5.62^2 -2 x 5.6 x 4.6 cos(π—∅_2) = 10.83 mm

Ym = 3.29 mm  

part (b)

sin(π-0.8 π)/3.29=sin(∅_R)/5.6

Solving for ∅_R we have  

∅_R=5.6*0.587/3.29

∅_R=85.5°=1.5 rad

Part(c)

If a third wave is also to be sent along the string in the same direction as the first two waves,then its phase angle must be in same as the resultant of the first 2 to maximize the amplitude.  

blue light is scattered in different directions by the tiny molecules of air. blue is scattered more than other colours because it has a shorter distance to travel which results in smaller waves. this is why we see a blue sky most of the time.