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STD 11 - 14. waves and sound — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 14. waves and sound4 Marks
MCQ
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
    $1.5$
  • ✓
    $1.07$
  • C
    $2.07$
  • D
    $0.5$
✓

Answer

Correct option: B.
$1.07$
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.$

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