Q115 / 200 · NEET 2004 · Physics

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

  • Correct answer
    A
    1.07 radians
  • B
    2.07 radians
  • C
    0.5 radians
  • D
    1.5 radians

Solution

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 $=$ $1.57-0.5=1.07$

Hence, the correct answer is option: 'A'.

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