The two waves are represented byy1 - 10-6 sin 100 t - x-50 - 0-5 my2 - 10-2 cos 100 t - x-50 mwhere x is in metres and t in seconds- The phase difference between the waves is approximately-

Y_1 = 10^ 6 sin ( 100 t + x/50 + 0.5 )m y_2 = 10^ 2 cos ( 100 t + x/50 )m ⇒ y_2 = 10^ 2 sin ( 100 t + x/50 + π/2 ) Phase difference = π/2 0.5 = 1.07 ra ...

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Standard XII

Physics

Phase Difference

The two waves...

Question

The two waves are represented by
$y_{1} = 10^{- 6} s i n (100 t + \frac{x}{50} + 0.5) m y_{2} = 10^{- 2} c o s (100 t + \frac{x}{50}) m$
where x is in metres and t in seconds. The phase difference between the waves is approximately:

1.07 rad

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2.07 rad

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0.5 rad

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1.5 rad

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Solution

The correct option is A 1.07 rad
$y_{1} = 10^{- 6} s i n (100 t + \frac{x}{50} + 0.5) m y_{2} = 10^{- 2} c o s (100 t + \frac{x}{50}) m \Rightarrow y_{2} = 10^{- 2} s i n (100 t + \frac{x}{50} + \frac{π}{2}) P h a s e d i f f e r e n c e = \frac{π}{2} - 0.5 = 1.07 r a d$

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Similar questions

Q. The phase difference between two waves represented by

y_{1} = 10^{- 6} sin [100 t + \frac{x}{50} + 0.5] m

y_{2} = 10^{- 6} cos [100 t + \frac{x}{50}] m

where x is expressed in metres and t is expressed in seconds is approximately

Q. The phase difference between two waves represented by

y_{1} = 10^{- 6} sin [100 t + (x / 50) + 0.5] m

y_{2} = 10^{- 6} cos [100 t + (x / 50)] m

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[Take

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Q. The phase difference between two waves represented by

y_{1} = 10^{- 6} sin [100 t + (x / 50) + 0.5] m

y_{2} = 10^{- 6} cos [100 t + (x / 50)] m

Where

x

is expressed in metre and

t

is expressed in second, is approximately
[Take

π = 3.14

]

Q. The two waves are represented by

y_{1} = 10^{- 6} s i n (100 t + \frac{x}{50} + 0.5) m y_{2} = 10^{- 2} c o s (100 t + \frac{x}{50}) m

where x is in metres and t in seconds. The phase difference between the waves is approximately:

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The two waves are represented by y1=10−6sin(100t+x50+0.5)my2=10−2cos(100t+x50)m where x is in metres and t in seconds. The phase difference between the waves is approximately:

The correct option is A 1.07 rady1=10−6sin(100t+x50+0.5)my2=10−2cos(100t+x50)m⇒y2=10−2sin(100t+x50+π2)Phase difference=π2−0.5=1.07 rad

The two waves are represented by
$y_{1} = 10^{- 6} s i n (100 t + \frac{x}{50} + 0.5) m y_{2} = 10^{- 2} c o s (100 t + \frac{x}{50}) m$
where x is in metres and t in seconds. The phase difference between the waves is approximately:

The correct option is A 1.07 rad
$y_{1} = 10^{- 6} s i n (100 t + \frac{x}{50} + 0.5) m y_{2} = 10^{- 2} c o s (100 t + \frac{x}{50}) m \Rightarrow y_{2} = 10^{- 2} s i n (100 t + \frac{x}{50} + \frac{π}{2}) P h a s e d i f f e r e n c e = \frac{π}{2} - 0.5 = 1.07 r a d$