Question

Wave of frequency 500 Hz has a phase velocity 360 m//s. The phase difference between two displacement at a certain point at time 10^(-3)s apart will be

Answer 1

Thus the distance between two points is 0.116 m and the phase difference is x = 0.35 m

Explanation:

Data given is as follow:

Velocity of the wave  

v  =  350  m  /  s

Frequency of the wave   n  =  500  H z

So wave length of the wave  

λ  =  v   / n  =  350  / 500  m  =  0.7  m

To find:

the distance between the two points which has  60 ∘

Out of phase i.e the phase difference is  

ϕ  =  60 ∘  =  π  / 3  r  a  d

Solution:

There is phase difference  2 π

we can say , if   ϕ  is the phase difference for path difference  x

then  ϕ  = 2 π x / λ

x  =  ϕ  λ  2  π  =  π  3  ×  0.7  2  π  =  0.7 6

m  ≈  0.116  m

2) Now in   t  =  10 ^− 3  s

The wave moves

v ×  t  =  350  ×  10 ^− 3  =  0.35  m

So here path difference  

x  =  0.35  m

Thus the distance between two points is 0.116 m and the phase difference is x = 0.35 m

Answer 2

$\huge\bold\red{HOLA!!}$

Given velocity of the wave v=350m/s. frequency of the wave n=500Hz ... 1) We are to find the distance between the two points which has 60∘ out of phase i.e the ... 2) Now in t=10−3s the wave moves