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The phase difference between two waves, represented by
$y_1 = 10^{-6}\sin\left\{100t + \dfrac{x}{50} + 0.5\right\}$ m
$y_2 = 10^{-6}\cos\left\{100t + \dfrac{x}{50}\right\}$ m
Where x is expressed in metres and t is expressed in seconds, is approximately : -
$y_1 = 10^{-6}\sin\left\{100t + \dfrac{x}{50} + 0.5\right\}$ m
$y_2 = 10^{-6}\cos\left\{100t + \dfrac{x}{50}\right\}$ m
Where x is expressed in metres and t is expressed in seconds, is approximately : -
A
2.07 radians
B
0.5 radians
C
1.5 radians
D
1.07 radians
Detailed Solution
To compare phases, write both waves as sine functions.
$y_1 = 10^{-6}\sin\left\{100t + \dfrac{x}{50} + 0.5\right\}$
$y_2 = 10^{-6}\cos\left\{100t + \dfrac{x}{50}\right\} = 10^{-6}\sin\left\{\dfrac{\pi}{2} + 100t + \dfrac{x}{50}\right\}$ (since $\cos\theta = \sin(\theta + \frac{\pi}{2})$)
Phase of $y_1$ $= 100t + \dfrac{x}{50} + 0.5$; phase of $y_2$ $= 100t + \dfrac{x}{50} + \dfrac{\pi}{2}$
Phase difference between $y_1$ and $y_2$ $= \dfrac{\pi}{2} - 0.5$
$= 1.57 - 0.5 \approx 1.07$ radians
So the phase difference is approximately 1.07 radians.
$y_1 = 10^{-6}\sin\left\{100t + \dfrac{x}{50} + 0.5\right\}$
$y_2 = 10^{-6}\cos\left\{100t + \dfrac{x}{50}\right\} = 10^{-6}\sin\left\{\dfrac{\pi}{2} + 100t + \dfrac{x}{50}\right\}$ (since $\cos\theta = \sin(\theta + \frac{\pi}{2})$)
Phase of $y_1$ $= 100t + \dfrac{x}{50} + 0.5$; phase of $y_2$ $= 100t + \dfrac{x}{50} + \dfrac{\pi}{2}$
Phase difference between $y_1$ and $y_2$ $= \dfrac{\pi}{2} - 0.5$
$= 1.57 - 0.5 \approx 1.07$ radians
So the phase difference is approximately 1.07 radians.