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The phase difference between two waves represented by
$\begin{aligned}
& y_1=10^{-6} \sin [100 t+(x / 50)+0.5] \mathrm{m} \\
& y_1=10^{-6} \cos [100 t+(x / 50)] \mathrm{m}
\end{aligned}$
where $x$ is expressed in metres and $t$ is expressed in seconds is approximately
Options:
$\begin{aligned}
& y_1=10^{-6} \sin [100 t+(x / 50)+0.5] \mathrm{m} \\
& y_1=10^{-6} \cos [100 t+(x / 50)] \mathrm{m}
\end{aligned}$
where $x$ is expressed in metres and $t$ is expressed in seconds is approximately
Solution:
1102 Upvotes
Verified Answer
The correct answer is:
1.07 radians
Here $y_1=10^6 \sin \left[100 t+\left(\frac{x}{50}\right)+0.5\right]$
$\begin{aligned}
& \text {and } y_2=10^{-6} \cos \left[100 t+\left(\frac{x}{50}\right)\right] \\
& =10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+\frac{\pi}{2}\right] \\
& =10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+1.57\right]
\end{aligned}$
Then comparing the equation of $y_1$ and $y_2$ we get phase difference $=$ $1.57-0.5=1.07$
$\begin{aligned}
& \text {and } y_2=10^{-6} \cos \left[100 t+\left(\frac{x}{50}\right)\right] \\
& =10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+\frac{\pi}{2}\right] \\
& =10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+1.57\right]
\end{aligned}$
Then comparing the equation of $y_1$ and $y_2$ we get phase difference $=$ $1.57-0.5=1.07$
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