The phase difference between two waves represented by {y_1} = {10^{ - 6}}sin [100,t + (x/50) + 0.5]m {y_2} = {10^{

Q: The phase difference between two waves represented by ${y_1} = {10^{ - 6}}\sin [100\,t + (x/50) + 0.5]m$ ${y_2} = {10^{ - 6}}\cos \,[100\,t + (x/50)]m$ where $ x$ is expressed in metres and $t$ is expressed in seconds, is approximately .... $ rad$

b (b) ${y_1} = {10^{ - 6}}\sin \,[100\,t + (x/50) + 0.5]$ ${y_2} = {10^{ - 6}}\sin \,\left[ {100\,t + \left( {\frac{x}{{50}}} \right) + \left( {\frac{\pi }{2}} \right)} \right]$ Phase difference $\phi$ $=[100 t + (x/50) + 1.57]-[100 t + (x/50)+ 0.5]$ $ = 1.07\,radians.$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

The phase difference between two waves represented by

${y_1} = {10^{ - 6}}\sin [100\,t + (x/50) + 0.5]m$

${y_2} = {10^{ - 6}}\cos \,[100\,t + (x/50)]m$

where $ x$ is expressed in metres and $t$ is expressed in seconds, is approximately .... $ rad$

A$1.5$
B$1.07$
C$2.07$
D$0.5$

AIPMT 2004, Medium

STD 11 Science
NEET
STD 11 - 14. waves and sound
Physics

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