How is Phase Difference of 4π Possible with AC = 7λ & BC = 5λ?

[Dexter Neutron]

Two different waves are starting froms points A,B respectively and going to a point C such that
$$AC - BC = 2\lambda$$
which corresponds to a phase difference of 4π
AC is given 7λ and BC is given 5λ

Now the second wave(BC) would need to have an initial phase of 4π so that the two waves can reach C at same time and interfere constructively
Thus Equations for wave AC,BC must be

$$y_1 = a\sin(\omega t + 4\pi)$$
since at t = 0 wave y1 is at 4π phase.
and
$$y_2 = a\sin(\omega t)$$

but in my book instead of +4π , -4π is written i.e.

$$y_1 = a\sin(\omega t - 4\pi)$$

How is that possible?

 

[sophiecentaur]

This is confusing and things appear, at first sight, to be the wrong way round. But:
AC is bigger, so C is further away. The wave arriving via AC will have left earlier than the wave via BC so the "-4π' corresponds to an Earlier time on the wave from A.
Likewise, the second part (-kx) in the equation for a progressive wave seems to be wrong, at first sight. The negative sign is there because the further away you go, the phase value is earlier in the history of the wave.
Thus:
y=a sin(ωt-kx)

 

[Dexter Neutron]

This is confusing and things appear, at first sight, to be the wrong way round. But:
AC is bigger, so C is further away. The wave arriving via AC will have left earlier than the wave via BC so the "-4π' corresponds to an Earlier time on the wave from A.
Likewise, the second part (-kx) in the equation for a progressive wave seems to be wrong, at first sight. The negative sign is there because the further away you go, the phase value is earlier in the history of the wave.
Thus:
y=a sin(ωt-kx)

For 4π is not kx but Φ the phase difference.
At t = 0 particle has displacement of 4π radians thus
y = asinθ = asin(4π)

and by the equation y = asin(wt+4π) we get on t = 0
y = asin(4π) thus the equation must be correct y = asin(wt+4π) and according to you we get y = asin(-4π) which is not true.

 

[nasu]

Actually
$a\sin(\omega t + 4\pi)$
is equal to
$a\sin(\omega t)$ (and also to $a\sin(\omega t - 4 \pi)$ )
To convince yourself, use the formula for the expansion of sin(a+b).

As you consider continuous waves and not finite pulses, phase differences which are multiple of 2π are irrelevant.
As the path difference is a multiple of the wavelength, the oscillations of the sources can be in phase or shifted by any multiple of 2π and they will still be in phase at point C.

 

[sophiecentaur]

For 4π is not kx but Φ the phase difference.

Firstly, it's just a number, added to the argument of sin(ωt). Why would the phase be different, if not because the wave has traveled further and arrives later and has a value due to what happened at an earlier time?

At t = 0 particle has displacement of 4π radians

That doesn't make sense. The displacement is a.sin(ωt-4π). The argument of a sin function is dimensionless.

y = asinθ = asin(4π)

+4π represents a wave that originated before t=0
4π is just a special case of a phase difference due to a path difference x. Apart from your personal version, can you find anywhere else where there is a positive sign for a wave traveling in the positive direction? Just Google "progressive wave" and get hundreds of hits.

 

[nasu]

The equations in OP are not "waves" but rather oscillations. He says that 4Pi would be initial phase. There is no position variable.
They may represent the source oscillations (the oscillations of points A and B) but not wave equations.
They may also represent the oscillation of the point C. The sign of the initial phase is not related to the propagation direction.
The wave propagating from A should be something like
y_A(x,t)= a sin(ωt - k x +∅_o)
where ∅_o is the initial phase.
The propagation direction is given by the sign in front of k and not by the sign of ∅_o.
For the one from B you may have
y_B(x,t)= a sin(ωt - k x)
If you want to have the two waves arriving in phase at a given point, situated at x_A from A and x_B from B you adjust the ∅o so that the two waves have the same phase

ωt - k x_A +∅_o =ωt - k x_B
(here x_A=AC and X_B=BC)
As xA>xB, the first wave (AC) needs to "start" earlier that the other one (BC).
So at the source, at t=0 and x=0, Y_B will be zero but y_A will be sin(∅o), so more advanced on the sine curve.
(For this ∅o must be positive.)But again, all this is irrelevant when you have continuous waves and path difference multiple of 2π. The waves will be in phase at C even if ∅o is zero, 2π or 100π.
Assuming that the waves go on forever. If you have a limited (in time) pulses of waves or continuous beams with limited coherence time then it will make a difference, of course. But it does not seem to be the case here.