Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.5k views
in Oscillations and waves by (20 points)
edited by

A plane transverse wave is propagating in a direction making an angle of \( 30^{\circ} \) with positive \( x \)-axis in the \( x \)-y plane. Find phase difference between points \( (0,0,0) \) and \( (1,1,1) \). Wavelength of the wave is \( 1 m \) :- 

(A) \( 2 \pi rad \) 

(B) \( (\sqrt{3}+1) \pi rad \) 

(C) \( (\sqrt{2}+1) \pi rad \) 

(D) None

Please log in or register to answer this question.

1 Answer

+2 votes
by (34.1k points)

OA = √2 cos 15° ⇒ △x

\(\because \) cos 15° = \(\frac{\sqrt3+1}{2\sqrt2}\)

Phase difference = \(\frac{2\pi}{\lambda}\) Path difference

△ϕ = \(\frac{2\pi}1\times\) △x

△ϕ = 2π x √2 cos15°

= 2π x √2 x \(\frac{\sqrt3+1}{2\sqrt2}\)

△ϕ = π (√3 + 1) rad

option (B)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...