The phase difference between two waves represented by aligned & y_1=10^ -6 [100 t+(x / 50)+0.5] m & y_1=10^ -6 [100 t+(x / 50)] m aligned where x is expressed in metres and t is expressed in seconds is approximately

Correct answer: A. 1.07 radians Here y_1=10^6 [100 t+((x)/(50))+0.5] aligned & and y_2=10^ -6 [100 t+((x)/(50))] & =10^ -6 [100 t+((x)/(50))+(π)/(2)] & =10^ -6 [100 t+((x)/(50))+1.57] aligned Then comparing the equation of y_1 and y_2 we get phase difference = 1.57-0.5=1.07

NEET 2004

Exam: NEET
Year: 2004
Subject: Physics
Chapter: Waves and Sound

NEET 2004 · Physics · Waves and Sound

The phase difference between two waves represented by aligned &...

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

A1.07 radians
B2.07 radians
C0.5 radians
D1.5 radians

Correct answer

A. 1.07 radians

Step-by-step 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$

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