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

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

2. Relevant equations
$$x = \frac{d\lambda}{a}$$

$$v = f\lambda$$

3. The attempt at a solution

$$f = 500Hz \ \ \ v = 340ms^{-1} \ \ \ a = 0.17 \times 10^{-2} cm$$

$$v = f\lambda$$

$$\lambda = \frac{v}{f} = \frac{340}{500} = 0.68m$$

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 Recognitions: Gold Member Science Advisor Staff Emeritus Two points 0.68cm apart will be out of phase by 360 degrees.

 Quote by Kurdt Two points 0.68cm apart will be out of phase by 360 degrees.
Two points $\lambda$ apart will be out of phase by 360 deg.

Regards,

Bill

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

In this case that is 0.68m. Damn I just realised I said cm, but you know what I mean.
 So, the answer is = 0.68m by 360° ?
 Recognitions: Homework Help 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 =]

 Quote by looi76 So, the answer is = 0.68m by 360° ?
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

Recognitions:
Gold Member
 $$0.68m = 360^o$$ $$0.17m = x$$ $$x = \frac{0.17 \times 360}{0.68}$$ $$x = 90^o$$ So, the phase difference is $$90^o$$ ?