I am not sure how to proceed or start for that matter 5:50 PM Thu Sep 1553. If U₁= sin (xx/L) cos (avt/L) and U₂ = sin (3xx/L) cos (3avt/L) are solutions to the waveequation, then show that U₁ + U₂ is a solution to the wave equation.U₁= Sin (TX) cos(Tvt)LU₂3 Sin (3x) cos (3x+)Tin (TX)U=U₁+U₂=Sin (TLE) COS(Txt) + Sin (31x) COS (3 TV+)L=-Sin/TTXTTV=Sin (ITE). ITX (- Sun (Txt)) + Sin (31x). 30+ (-5m (2xx²))LL@63%Wove cq.: 1Nave+ : 0J²U (x, t) = | J²U (x, t)dx²²¹39+

Answered: Image /qna-images/answer/2159bb4b-6f57-4dc3-aba8-e2678ee50792.jpg

< 5 IMMAN SANA." 3. If U₁ = sin (xx/L) cos (avt/L) and U₂ = sin (3xx/L) cos (3avt/L) are solutions to the wave equation, then show that U₁ + U₂ is a solution to the wave equation. U.: Sin (TA) cos(Tvt) U₂3 Sin (31x) cos (³x+) Sin (TX) TO =-Sin U=U₁+U₂ = Sin (TX) COS (Txt) + Sin (3πx) COS (³ TV+) = TTX Sin (ITE). ITX (- Sun (Txt)) + Sin (31x). 30* (- 5m (2+x²)) -vt L L HO Nove cq. J'U (xt). I d'Ult (x, dx² +: 0

< 5 IMMAN SANA." 3. If U₁ = sin (xx/L) cos (avt/L) and U₂ = sin (3xx/L) cos (3avt/L) are solutions to the wave equation, then show that U₁ + U₂ is a solution to the wave equation. U.: Sin (TA) cos(Tvt) U₂3 Sin (31x) cos (³x+) Sin (TX) TO =-Sin U=U₁+U₂ = Sin (TX) COS (Txt) + Sin (3πx) COS (³ TV+) = TTX Sin (ITE). ITX (- Sun (Txt)) + Sin (31x). 30* (- 5m (2+x²)) -vt L L HO Nove cq. J'U (xt). I d'Ult (x, dx² +: 0

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter9: Atomic Absorption And Atomic Fluorescence Spectrometry

Section: Chapter Questions

Problem 9.15QAP

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