Question

A wave of frequency 500 Hz has a wave velocity of 350 m//s. (a) Find the distance between two points which are 60@ out of phase. (b) Find the phase difference between two displacement at a certain point at time 10^(-3) s apart.

Answer 1

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

Explanation:

Given data:

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

(1) We are 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

As we know that for path difference  λ

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

Given :

The Velocity of the wave = 350 m/s

The Frequency of the wave = 500 H z

To Find :

(a) The distance between two points, 60° out of phase

(b) The  phase difference between two displacement at a certain point at time $10^{-3}$ s apart

Solution :

∵  wave length of the wave = $\dfrac{velocity}{frequency}$

                                             = $\dfrac{350 m/s}{500 Hz}$

                                             = 0.7  meter

(a)

∵  ∅ = 60°

And   180° = π radian  

So,   60° = $\dfrac{\pi }{180}$ × 60°   =  $\dfrac{\pi }{3}$  rad

i.e phase difference =   ∅  = $\dfrac{\pi }{3}$  rad

Since , For path difference of $\lambda$ , we have phase difference of 2π ,

So, for ∅ phase difference, we have path difference ∅ = 2π × $\dfrac{x}{\lambda }$

i.e  ∅ = 2π × $\dfrac{x}{\lambda }$

Or, $\dfrac{\pi }{3}$  = 2 × π × $\dfrac{x}{0.7}$

Or,  x = $\dfrac{0.7}{6}$

∴    x = 0.1167  meters

So,  The distance between two points =  x = 0.1167  meters

( b )

∵  Time = $10^{-3}$ sec

From wave motion , we can write

Displacement = velocity × time

                      = 350 m/s × $10^{-3}$ sec

                      = 0.35 meters

Again

∵ Phase difference =  ∅ = 2π × $\dfrac{x}{\lambda }$

So,      ∅ = 2 × π × $\dfrac{0.35 m}{0.7 m }$

Or,      ∅ = 2 × π × 0.5

∴        ∅ = π

So, Th value of phase difference = ∅ = π

Hence,

(a )  The distance between two points is 0.1167  meters

(b ) Th value of phase difference between two displacements is π  Answer