Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • Two waves <img alt="\mathrm{y}_{1}=\mathrm{A}_{1} \sin \left(\omega \mathrm{t}-\beta_{1}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%7D_%7B1%7D%3D%5Cmathrm%7BA%7D_%7B

Two waves \mathrm{y}_{1}=\mathrm{A}_{1} \sin \left(\omega \mathrm{t}-\beta_{1}\right) and \mathrm{y}_{2}=\mathrm{A}_{2} \sin \left(\omega \mathrm{t}-\beta_{2}\right) superimpose to form a resultant wave. The amplitude of this resultant wave is given by

Option: 1

\mathrm{A_{1}+A_{2}}


Option: 2

\mathrm{\left|A_{1}-A_{2}\right|}


Option: 3

\mathrm{\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \left(\beta_{1}-\beta_{2}\right)}}


Option: 4

\mathrm{\sqrt{A_{1}^{2}+A_{2}^{2}-2 A_{1} A_{2} \cos \left(\beta_{1}-\beta_{2}\right)}}


Answers (1)

best_answer

\mathrm{Phase\: diff =\omega \mathrm{t}-\beta_{1}-\omega \mathrm{t}+\beta_{2}=\beta_{2}-\beta_{1}}

\mathrm{\therefore }  Amplitude of resultant wave

\mathrm{=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \left(\beta_{2}-\beta_{1}\right)} }
\mathrm{=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \left(\beta_{1}-\beta_{2}\right)} \quad[\text { As } \cos (-\theta)=+\cos \theta] }

Posted by

admin

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

What is an APAAR ID? NTA mandates Aadhaar verification to enhance integrity of NEET UG 2025
anam-khan

NTA asks NEET UG 2025 aspirants to use APAAR ID, update Aadhar details for registration
anam-khan

NEET 2025 exam in hybrid mode and two phases? What will be different this year

Latest Question