[SOLVED] What is the phase difference between the motions of two points on the wave..

looi76

1. The problem statement, all variables and given/known data
The frequency of a certain wave is [tex]500Hz[/tex] and its speed is [tex]340ms^{-1}[/tex].What is the phase difference between the motions of two points on the wave [tex]0.17cm[/tex] apart?

2. Relevant equations
[tex]x = \frac{d\lambda}{a}[/tex]

[tex]v = f\lambda[/tex]

3. The attempt at a solution

[tex]f = 500Hz \ \ \ v = 340ms^{-1} \ \ \ a = 0.17 \times 10^{-2} cm[/tex]

[tex]v = f\lambda[/tex]

[tex]\lambda = \frac{v}{f} = \frac{340}{500} = 0.68m[/tex]

What should I do next?

Kurdt

Two points 0.68cm apart will be out of phase by 360 degrees.

Antenna Guy

Two points [itex]\lambda[/itex] apart will be out of phase by 360 deg.

Regards,

Bill

Kurdt

In this case that is 0.68m. Damn I just realised I said cm, but you know what I mean.

looi76

So, the answer is = 0.68m by 360° ?

Gib Z

No, I won't be. You know the phase difference between points 0.68m apart is 360°, the question asks the phase difference for 0.17m, which happens to be a quarter the quantity of which we do know the phase difference for. Just one more step =]

Antenna Guy

Define the terms in your first relevant equation - one of those terms should be what you solve for to get the answer.

BTW - check what you wrote for "a" (cm?).

Regards,

Bill

Kurdt

As GibZ alluded to, I was trying to coax you toward the right answer not just give you the answer outright (since that would be against forum policy).

looi76

[tex]0.68m = 360^o[/tex]
[tex]0.17m = x[/tex]

[tex]x = \frac{0.17 \times 360}{0.68}[/tex]

[tex]x = 90^o[/tex]

So, the phase difference is [tex]90^o[/tex] ?

Kurdt

Thats correct.