Question

Two waves are represented by the equations y1= asin(wt+kx+0.57)m and y2=acos(wt+kx)m where x is in metre and t in second. The phase difference between them is?

1- 0.57 rad 2- 1.0 rad

3-1.25rad 4- 1.57 rad

Answer 1

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Answer 2

Answer:

Two waves represented by y1= a sin(wt+kx+0.57)m and y2=a cos(wt+kx)m. The phase difference between them is 1 radian.

Explanation:

• A harmonic wave travelling towards positive x direction, i.e. from left to right is given by its displacement equation

$y = a \sin(kx - wt +∅₀ ) - - - (1)$

• The argument of the sine function, i.e. ( kx - wt + ∅₀ ) is called the phase ∅ and ∅₀ is called the phase angle or phase constant.
• Thus, phase at any position x and at any instant t in the harmonic wave represented by equation ( 1 ) is given by

$∅ = (kx - wt + ∅₀)$

• The phase difference between two waves represented by

y = a sin ∅1 and y = a sin ∅2 is

• ∆∅ = ∅1 - ∅2.

Given that:

y1 = a sin(wt+kx+0.57) and y2 =a cos(wt+kx)

Solution :

• Two waves are represented by:

$y1 = a \sin(wt + kx + 0.57) \\ y2 = a \cos(wt + kx) \\ y2 = a \sin( \frac{\pi}{2} + wt + kx )$

• Comparing both given equation of waves with equation (1), we get:

$∅1 = wt + kx + 0.57 \\ ∅2 = \frac{\pi}{2} + wt + kx$

• The phase difference is given by:
• ∆∅ = ∅1 - ∅2

$∆∅ = (wt + kx + \frac{\pi}{2} ) - (wt + kx + 0.57) \\ = \frac{\pi}{2} - 0.57 \\ = 1.57 - 0.57 \\ = 1 \: radian$

• Hence, the phase difference between given two waves is 1 radian.