Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 sec and the velocity of the wave is 300 m/sec. What is the phase difference between the oscillations of two points?
Answers
Question :
Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 s and the velocity of the wave is 300 m/s. What is the phase difference between the oscillation of two points?
Theory :
pase diffrence :
At any instant t, if Φ1 and Φ2 are the phases of two particles with distances from the origin x1 and x2 . Then ,
Φ1 = (wt-kx1)
and
Φ2 = (wt-kx2)
⇒ Φ1- Φ2 = k(x2- x1)
Given:
Time period ,T = 0.05 sec
velocity of wave ,v = 300 m/s
frequency , f =
⇒
⇒f = 20Hz
Δ x = 15-10 = 5 m
we know that
λ = v/f
put the value ofλ and Δx
⇒
Answer :
The phase difference between the oscillations of two points is 2.093
Explanation:
Given path difference = 15 m - 10 m = 5 m
The time period of oscillation T= 0.05 s.
Frequency,
Velocity of wave v= 300 m/s
So wavelength,
To calculate the phase difference use relation between path and phase difference,
Phase difference × path difference
Substitute the given values, we get
Phase difference
Phase difference
Thus, phase difference is 2.093.