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

[utkarshakash]

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??

 

[BruceW]

You also need to take time into account. The phase will be [ (2π/λ) path difference ] plus some time dependence.

 

[utkarshakash]

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?

 

[Tanya Sharma]

Hi utkarshakash...

What's the answer ? Is it 7sec ?

 

[ehild]

What function of distance and time is called wave?

ehild

 

[Tanya Sharma]

Hello ehild...

Do you get 7sec as the answer ?

 

[ehild]

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

 

[utkarshakash]

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 $y=f(t-\frac{x}{v})$. If I do follow your approach, then I end up with this:

$kx-\omega t = \pi \\<br /> \frac{2\pi}{\lambda} (5-t) = \pi \\<br /> t=3$

 

[utkarshakash]

Hi utkarshakash...

What's the answer ? Is it 7sec ?

Yes.

 

[ehild]

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

 

[nasu]

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.