How to calculate phase difference for spherical waves?

how to calculate phase difference for spherical waves?how to say whether they are in phase or out of phase?
in sinusoidal we can easily say whether they are in phase or out of phase just by looking at it,but how to do the same for spherical waves?

Just assume that they are sinusoidal spherical waves.

Just assume that they are sinusoidal spherical waves.
would be gratified if you could give an image which depicts that.

How about this one? These are circular rather than spherical. The waves in the water you mentioned are not actually "spherical" anyway.
http://matterwaves.info/images/proton08.gif
thank you very much for providing that image,sir,yes ripples are circular,not spherical.sorry for that.but can you please tell me what do those waves represent?(means what is y axis and x axis)

blue_leaf77
Homework Helper
When you have two spherical waves and want to find the phase difference between the waves observed at some point, you have to specify the position of both sources and the position of observer. Suppose there two point sources in 3D Cartesian coordinate at ##\mathbf{r}_1## and ##\mathbf{r}_2##, and the observer stands in the point ##\mathbf{r}_o##. The total disturbance at the observer will be
$$U(\mathbf{r}_o) = A_1\frac{e^{ik|\mathbf{r}_1-\mathbf{r}_o|}}{|\mathbf{r}_1-\mathbf{r}_o|} + A_2\frac{e^{ik|\mathbf{r}_2-\mathbf{r}_o|}}{|\mathbf{r}_2-\mathbf{r}_o|}$$
with ##A_1## and ##A_2## the amplitude of the sources and ##k=2\pi/\lambda## the wavenumber. The phase difference felt by the observer will then be
$$\Delta \Phi (\mathbf{r}_o) = k\left( |\mathbf{r}_1-\mathbf{r}_o| - |\mathbf{r}_2-\mathbf{r}_o| \right).$$

When you have two spherical waves and want to find the phase difference between the waves observed at some point, you have to specify the position of both sources and the position of observer. Suppose there two point sources in 3D Cartesian coordinate at ##\mathbf{r}_1## and ##\mathbf{r}_2##, and the observer stands in the point ##\mathbf{r}_o##. The total disturbance at the observer will be
$$U(\mathbf{r}_o) = A_1\frac{e^{ik|\mathbf{r}_1-\mathbf{r}_o|}}{|\mathbf{r}_1-\mathbf{r}_o|} + A_2\frac{e^{ik|\mathbf{r}_2-\mathbf{r}_o|}}{|\mathbf{r}_2-\mathbf{r}_o|}$$
with ##A_1## and ##A_2## the amplitude of the sources and ##k=2\pi/\lambda## the wavenumber. The phase difference felt by the observer will then be
$$\Delta \Phi (\mathbf{r}_o) = k\left( |\mathbf{r}_1-\mathbf{r}_o| - |\mathbf{r}_2-\mathbf{r}_o| \right).$$
sir,this equation is much beyond my portion,basically i did not wish to find the numerical value.what i wanted to know is,when we take two sinusoidal waves and just by looking at it we can say whether they are in phase or out of phase,(by superimosing one on another),how to do the same for spherical waves since there is no crest or trough,how can we superimpose?

blue_leaf77
Homework Helper
Let me rewrite the total disturbance in terms of sine functions
$$U(\mathbf{r}_o) = A_1\frac{\sin(k|\mathbf{r}_1-\mathbf{r}_o|)}{|\mathbf{r}_1-\mathbf{r}_o|} + A_2\frac{\sin(k|\mathbf{r}_2-\mathbf{r}_o|)}{|\mathbf{r}_2-\mathbf{r}_o|}$$
when we take two sinusoidal waves and just by looking at it we can say whether they are in phase or out of phase,(by superimosing one on another),how to do the same for spherical waves since there is no crest or trough,how can we superimpose?
Spherical or circular waves are a bit different from purely sinusoidal waves. When you said sinusoidal I imagine what you are picturing in your mind is a sinusoidal disturbance in two dimensions. Taking an example of string vibration, you only need two dimension, x and y, to draw it. A different method of drawing is to be expected when, instead, you want to draw circular wave, like ripple on a water surface. You need three dimensions to visualize such disturbance because the wave spreads in a plane (make it the xy plane), and the disturbance occurs in a direction perpendicular to the water surface (make it the z direction). One more dimension will be needed to draw spherical wave in space.

Furthermore, you cannot apply a direct analogy in 1D sinusoidal wave into spherical/circular wave, because 1D wave assumes that the wave disturbance is homogenous everywhere and thus you don't need to specify the position of the source. For circular/spherical waves, the fact that the wave spread radially means that the relative position of the source and observer matters.

how to do the same for spherical waves since there is no crest or trough,how can we superimpose?
By definition, waves are periodic disturbances. Therefore one can always identify the crest and through, see the equation above, there are sine terms which obviously have crests and throughs. The way to superimpose them is by calculating that equation above. Unlike 1D waves, it's indeed impossible to find the closed form of the resultant disturbance for circular/spherical waves.

Let me rewrite the total disturbance in terms of sine functions
$$U(\mathbf{r}_o) = A_1\frac{\sin(k|\mathbf{r}_1-\mathbf{r}_o|)}{|\mathbf{r}_1-\mathbf{r}_o|} + A_2\frac{\sin(k|\mathbf{r}_2-\mathbf{r}_o|)}{|\mathbf{r}_2-\mathbf{r}_o|}$$

Spherical or circular waves are a bit different from purely sinusoidal waves. When you said sinusoidal I imagine what you are picturing in your mind is a sinusoidal disturbance in two dimensions. Taking an example of string vibration, you only need two dimension, x and y, to draw it. A different method of drawing is to be expected when, instead, you want to draw circular wave, like ripple on a water surface. You need three dimensions to visualize such disturbance because the wave spreads in a plane (make it the xy plane), and the disturbance occurs in a direction perpendicular to the water surface (make it the z direction). One more dimension will be needed to draw spherical wave in space.

Furthermore, you cannot apply a direct analogy in 1D sinusoidal wave into spherical/circular wave, because 1D wave assumes that the wave disturbance is homogenous everywhere and thus you don't need to specify the position of the source. For circular/spherical waves, the fact that the wave spread radially means that the relative position of the source and observer matters.

By definition, waves are periodic disturbances. Therefore one can always identify the crest and through, see the equation above, there are sine terms which obviously have crests and throughs. The way to superimpose them is by calculating that equation above. Unlike 1D waves, it's indeed impossible to find the closed form of the resultant disturbance for circular/spherical waves.
sir, your explanations are very advanced,iam still in high school.please provide an explanation which is easy to understand.

thank you very much for providing that image,sir,yes ripples are circular,not spherical.sorry for that.but can you please tell me what do those waves represent?(means what is y axis and x axis)
The picture shows the maxima and minima of the waves as different shades of grey.
The graph bellow shows something like a cross section of the wave at a given time. The horizontal axis represent distance along one radius (starting from the centere of the circle. The vertical axis is the displacement at at each point along that radius, at a given time.

You should realize that the "sinusoidal" character of a wave is independent of the shape of the wave-fronts or of the polarization (longitudinal or transverse).
A wave is sinusoidal if the equation describing the wave can be written in terms of sines (and cosines). Even if the wave is not a pure sine waves, it can be written as a sum of sine waves.
For example, a sine wave can be
y=A sin(wt - kx)
Where A is the amplitude, w is the frequency and k is the wave-number.
y could be the vertical displacement for a transverse wave on a string but it can also be the longitudinal displacement for a longitudinal waves on a slinky. Or it can be pressure for a sound wave in air (also longitudinal).
For a plane wave A (which also can be displacement in various directions or pressure or electric field) is constant, same value at any point.
For circular or spherical waves A is a function of position. But the "sin" is still there.
Sure, waves can be more complicated than just a pure sine wave but this has nothing to do with them being circular rather than plane waves.

blue_leaf77
\begin{aligned} |\mathbf{r}_1-\mathbf{r}_o| &= \sqrt{(x_o-5)^2+y_o^2} \\ |\mathbf{r}_2-\mathbf{r}_o| &= \sqrt{(x_o+5)^2+y_o^2} \\ \end{aligned}