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< 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

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter5: Electron Configurations And The Periodic Table
Section: Chapter Questions
Problem 26QRT
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5:50 PM Thu Sep 15 5 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. U₁= Sin (TX) cos(Tvt) L U₂3 Sin (3x) cos (3x+) T in (TX) U=U₁+U₂ = Sin (TLE) COS(Txt) + Sin (31x) COS (3 TV+) L =-Sin/TTX TTV = Sin (ITE). ITX (- Sun (Txt)) + Sin (31x). 30+ (-5m (2xx²)) L L @63% Wove cq.: 1 Nave + : 0 J²U (x, t) = | J²U (x, t) dx² ²¹ 3 9 +
Transcribed Image Text:5:50 PM Thu Sep 15 5 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. U₁= Sin (TX) cos(Tvt) L U₂3 Sin (3x) cos (3x+) T in (TX) U=U₁+U₂ = Sin (TLE) COS(Txt) + Sin (31x) COS (3 TV+) L =-Sin/TTX TTV = Sin (ITE). ITX (- Sun (Txt)) + Sin (31x). 30+ (-5m (2xx²)) L L @63% Wove cq.: 1 Nave + : 0 J²U (x, t) = | J²U (x, t) dx² ²¹ 3 9 +
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