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The wave velocity of a progressive wave is 480ms1and the phase difference between the two particles separated by a distance of 12m is 10800. The number of waves passing across a point in 1 sec is
A. 120
B. 240
C. 60
D. 360

Answer
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Hint:To solve this question you have to use the relation between the phase difference and path difference. The phase difference is defined as the difference in the phase angle of the two waves and the Path difference is defined as the difference in the path travelled by the two waves. Hence there is a direct relation between phase difference and path difference is. Both are directly proportional to each other.

Formula used:
In any two waves with the same frequency, the relation between Phase Difference and Path Difference is given as -
Δϕ=2πλΔx
Where Δx is the path difference between the two waves and Δϕ is the phase difference between the two waves.

Complete step by step solution:
Given: Phase difference, Δϕ=1080180π=6π
Wave velocity, v=480ms1
Separation distance, Δx=12m
As we know that
Δϕ=2πλΔx
λ=2πΔϕΔx
Substituting the values, we have
λ=2π6π×12
λ=4m
Now the number of waves passing can be,
n=vλ
On substituting the values,
n=4804
n=120
Therefore, the number of waves passing across a point in 1 sec is 120.

Hence option A is the correct answer.

Note: There is a direct relation between Phase Difference and Path as they are directly proportional to each other. Phase difference is the difference between phase angles between two waves. On the other hand, Path difference refers to the difference in the path travelled by the two waves.
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