Skip to main content
Back to WAVES

JEE PHYSICS MCQ: WAVES

1 Medium
1 of 1
Analyzing +4 | -1
The phase difference between the two given waves is y1=a1sin⁡(ωt−2πλX)   y2=a2cos⁡(ωt−2πλX+ϕ) {y_1} = {a_1}\sin \left( {\omega t - \frac{{2\pi }}{\lambda }X} \right)\,\,\,{y_2} = {a_2}\cos \left( {\omega t - \frac{{2\pi }}{\lambda }X + \phi } \right)\,
View answer and explanation
Correct option: C
π2+(ϕ)\frac{\pi }{2} + \left( \phi \right)
Option C (π2+(ϕ)\frac{\pi}{2} + (\phi)) is correct because to find the phase difference, both waves must be expressed as sine functions, and then the difference in their arguments is calculated.

More Questions from WAVES

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.