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

Suggest Corrections

Similar questions

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

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

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

y_{1} = a s i n (ω t + k x + 0.57) m

y_{2} = a c o s (ω t + k x) m

x

t

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

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

x

t

π = 3.14

y_{1} = a s i n (ω t - x)

y_{2} = b c o s (ω t - x)

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