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.  

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.