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`        The pharse  between difference between the two ways represent byy1=10^-6sin[100t+x/50+0.5]my2=10^-6cos[100t+x/50]mwhere x is expressed in metres and t is expressed in sec is approximately`
10 months ago

Arun
22615 Points
```							The phase difference can be observed if both the expressions are in either cosin or sin . so after converting both in cosin or sin, one can compare both expressions and the difference between the phase of two could be find out. hence y1 = 10^-6 cos (pi/2 – 0.5 – (100t + x/50)) hence phase difference = pi/2 – 0.5 = 1.08 radian approximately
```
10 months ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions