the frequency of a certain wave is 500Hz and its speed is 340 ms^-1
what is the phase difference between the motions of the two points on the wave 0.17m apart?

Question

the frequency of a certain wave is 500Hz and its speed is 340 ms^-1 
what is the phase difference between the motions of the two points on the wave 0.17m apart?

kropot72 · Answer

$$wavelength=\frac{speed}{frequency}=\frac{340}{500}\ m$$
One wavelength occupies one cycle. One cycle takes 360 degrees.
Therefore the required phase difference is given by
$$360\times \frac{0.17}{wavelength}$$

anonymous · Answer

the answer is pie upon  2 rad . how come?

kropot72 · Answer

$$\frac{\pi}{2}\ radians=90\ degrees$$

anonymous · Answer

Do you need help solving the entire thing?

kropot72 · Answer

If the answer is required in radians the solution is given by
$$2 \pi \times \frac{0.17}{0.68}$$

anonymous · Answer

@blackilluminati  yes

anonymous · Answer

frequency= 500hz 
Speed= 340ms-1 
Relevant equations x= d x wavelength/ a 
Wavelength= Speed/frequency 
340/500= 0.68m 
Two points will be out of phase by 360* (*=Degree) 
0.68m= 360* 
therefore what is the angle for 0.17m=? 
You cross multiply 0.17 x 360/ 0.68= 90*
There you go (: