Phase Angle of Point 5m from Wave Source: 180° at 2π Seconds

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In summary, the speed of the wave is 1m/s along a straight line and the initial phase angle is 0. The wavelength is 4m and the point 5m from the wave source will have a phase of 180° after a total time of 5 seconds plus half a period.
  • #1
utkarshakash
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Homework Statement


A wave travels along a straight line with speed 1m/s . The initial phase angle is 0. Wavelength = 4m. Measured from initiation of waves, when will the phase of the point 5m from the wave source first be 180°?



The Attempt at a Solution



Phase difference = (2π/λ) path difference

π = (2π/λ) 5.

But I really have no idea what I'm doing here. The LHS and RHS are not equal. What's this??:confused:
 
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  • #2
You also need to take time into account. The phase will be [ (2π/λ) path difference ] plus some time dependence.
 
  • #3
BruceW said:
You also need to take time into account. The phase will be [ (2π/λ) path difference ] plus some time dependence.

Sorry but I still can't see what should be added to it. Can you please give some more hints?
 
  • #4
Hi utkarshakash...

What's the answer ? Is it 7sec ?
 
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  • #5
What function of distance and time is called wave?

ehild
 
  • #6
Hello ehild...

Do you get 7sec as the answer ?
 
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  • #7
I can not tell the answer, sorry. Start with the function that describes a sinusoidal traveling wave, y(x,t) = Asin(kx-wt), which is zero at t=0 and x=0. The phase is kx-wt. When will be first the phase pi at x=5?

ehild
 
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  • #8
ehild said:
I can not tell the answer, sorry. Start with the function that describes a sinusoidal traveling wave, y(x,t) = Asin(kx-wt), which is zero at t=0 and x=0. The phase is kx-wt. When will be first the phase pi at x=5?

ehild

The question does not mention anywhere that the wave is sinusoidal. I think a more general approach would be to assume the function as [itex]y=f(t-\frac{x}{v}) [/itex]. If I do follow your approach, then I end up with this:

[itex]kx-\omega t = \pi \\
\frac{2\pi}{\lambda} (5-t) = \pi \\
t=3 [/itex]
 
  • #9
Tanya Sharma said:
Hi utkarshakash...

What's the answer ? Is it 7sec ?

Yes.
 
  • #10
You are right, if you want the phase increase with time at a given point you should choose the wave as f(t-x/v), instead of f(x-vt). But the form of the phase was not defined, and they are of opposite signs. That is a badly-worded problem, again.

ehild
 
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  • #11
You don't need to assume a specific form. Just that the phase changes by 360 in one period.
The wave needs 5s to reach the point 5 m away from the source. When it does, it will have a phase of zero, as the wave front with phase zero just reached this point. For the phase at this point to change from zero to 180 you need to wait another half period.
So it is 5s + T/2.
 

FAQ: Phase Angle of Point 5m from Wave Source: 180° at 2π Seconds

1. What does the term "phase angle" refer to in this context?

The phase angle refers to the relative position of a point on a wave in relation to a reference point. In this case, it is the position of a point 5m from the wave source.

2. How is the phase angle of a point 5m from a wave source measured?

The phase angle is typically measured in degrees or radians, and is calculated by dividing the distance of the point from the wave source by the wavelength of the wave.

3. Why is the phase angle of a point 5m from a wave source important?

The phase angle can provide information about the properties of a wave, such as its frequency and wavelength. It is also important in understanding interference patterns and determining the relative positions of different points on a wave.

4. What does it mean for the phase angle to be 180° at 2π seconds?

A phase angle of 180° means that the point is at its maximum displacement from the reference point, or is at the crest or trough of the wave. The 2π seconds refers to the time it takes for the wave to complete one full cycle.

5. How does the phase angle change as the distance from the wave source increases?

As the distance from the wave source increases, the phase angle will also change. This is because the wavelength of the wave will also increase, causing the point to complete more cycles in a given time period. This results in a larger phase angle.

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