The phase difference between two sound waves represented by $y_{1} = A_{1} sin (w t - k x)$ and $y_{2} = A_{2} cos (w t - k x + π / 3)$ is given by

π / 3

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π / 6

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π / 4

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π / 2

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Solution

The correct option is A $π / 3$
Phase difference is the difference in phases = $π / 3$
The correct option is (a)

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Similar questions

Two waves are given by $y_{1} = a s i n (ω t - k x)$ and $y_{2} = a c o s (ω t - k x)$ The phase difference between the two waves is

[MP PMT 1993; SCRA 1996; CET 1998;EAMCET 1991; Orissa JEE 2002]

y_{1} = 0.1 s i n (100 π t + \frac{π}{3})

y_{2} = 0.1 c o s 100 π t

y_{1} = A sin (ω t - k x)

y_{2} = A sin (ω t - k x + \frac{π}{4})

y_{1} = 8 sin (ω t + k x); y_{2} = 6 sin (ω t + \frac{π}{2} + k x); y_{3} = 4 sin (ω t + π + k x);

y_{4} = 2 sin (ω t + \frac{3 π}{2} + k x);

y_{5} = 4 \sqrt{2} sin (ω t - k x + \frac{π}{4})

y_{1} = 0.1 sin (100 π t + \frac{π}{3})

y_{2} = 0.1 cos π t

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