What is the phase difference for waves with a 25% increase in amplitude?

[RyRy19]

Two waves are identical apart from a phase difference, they create a resultant wave of 25% increase of the original amplitude, what is the phase difference.
sin(alpha)+sin(beta)=2sin(1/2)(alpha+beta)cos(1/2)(alpha-beta)
Asin(kx+wt)=2sin(1/2)(alpha+beta)cos(1/2)alpha-beta)
Asin(kx+wt)=2sin(1/2)(alpha+beta)cos(1/2)alpha-beta)
=1.2Asin(kx+wt)

I assume that's right then you rearrange but I'm highly doubting myself

 

[Simon Bridge]

It is a LOT easier if you use the phasor representation of the waves.
Then you just add the phasor arrows head-to-tail like vectors.

 

[RyRy19]

It is a LOT easier if you use the phasor representation of the waves.
Then you just add the phasor arrows head-to-tail like vectors.

So like.. e^i(wt-kx) ? I never really understood this method so I should probably look it up, how would you apply that to this question? If I may ask

Thanks for the help!

 

[Simon Bridge]

The two waves vary at the same rate - so they will always maintain the same phase difference between them.

Executive summary:

Waves can be represented by vectors that rotate. The length of the vector is the amplitude of the wave.

Two waves that are identical but for a phase difference are two vectors with an constant angle between them.

The result of the two waves combining is the same as adding the two vectors.

So two waves with a phase difference of $\phi$ but identical otherwise, will form a iscoseles triangle where the external angle at the apex is the phase difference. Thus the internal angle is $\theta=\pi-\phi$

The resultant is the third side - use the cosine rule.
Sketch it and you'll see.

i.e. if the phase difference is $\pi/2$, and the amplitudes are A, then the resultant wave will have amplitude $\sqrt{2}A$ (pythagoras).

What you are trying to do is:
$\sin(\omega t-kx)+\sin(\omega t-kx+\phi) = A\sin(\omega t-kx+\delta)$
... you are given $A$ and you need to find $\phi$

 

[RyRy19]

So to ask simply, apologies if I'm wrong. As we know that both waves have the same amplitude and their resultant is 25% greater, could I draw a line joined to another line which as at an angle. Label the angle outside of the two lines (apex) as theta. After this I could use Pythagoras to determine the inside angles and then knowing the angle is 180 determine the unknown angle which is he phase difference?

I could label lines with length 4, 4 and 5
So sorry for the lateness!

 

[Simon Bridge]

I'd draw arrows, not lines. In this case they are the same length.
As vectors: $\vec r = (r,\theta)$ ... i.e. a length and an angle measured anticlockwise from some reference direction (i.e. the x axis).

If the first wave is $\vec r_1 = (1,0)$
Then the second wave is $\vec r_2 = (1,\theta)$
Sketch these head-to-tail - careful: $\theta\neq\pi/2$ so don't draw a right-angle.

Since there is no right angle, you cannot use pythagoras directly.
I'd use the cosine rule instead.

 

[Simon Bridge]

I'd draw arrows, not lines. In this case they are the same length.
As vectors: $\vec r = (r,\theta)$ ... i.e. a length and an angle measured anticlockwise from some reference direction (i.e. the x axis).

If the first wave is $\vec r_1 = (1,0)$
Then the second wave is $\vec r_2 = (1,\theta)$
Sketch these head-to-tail - careful: $\theta\neq\pi/2$ so don't draw a right-angle.

Since there is no right angle, you cannot use pythagoras directly.
I'd use the cosine rule instead.