Lecture 8 Notes

# Lecture 8 Notes

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t) + ω0 B cos(ω0 t)) x￿￿ (t) = −ω0 A sin(ω0 t) + ω0 B cos(ω0 t) p +(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) 2 2 +t(−ω0 A cos(ω0 t) − ω0 B sin(ω0 t)) Forced vibrations (3.8) • Case 2: ω = ω0 F0 ω0 = x+ = cos(ω0 t) m xp (t) = t(A cos(ω0 t) + B sin(ω0 t)) ￿￿ ￿ xp (t) 2 ω0 x ￿ k m = A cos(ω0 t) + B sin(ω0 t) +t(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) x￿￿ (t) = −ω0 A sin(ω0 t) + ω0 B cos(ω0 t) p +(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) 2 2 +t(−ω0 A cos(ω0 t) − ω0 B sin(ω0 t)) Forced vibrations (3.8) ω = ω0 • Case 2: F0 ω0 = x+ = cos(ω0 t) m xp (t) = t(A cos(ω0 t) + B sin(ω0 t)) ￿￿ ￿ xp (t) 2 ω0 x ￿ k m = A cos(ω0 t) + B sin(ω0 t) +t(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) x￿￿ (t) = −ω0 A sin(ω0 t) + ω0 B cos(ω0 t) p +(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) A=0 2 2 +t(−ω0 A cos(ω0 t) − ω0 B sin(ω0 t)) Forced vibrations (3.8) ω = ω0 • Case 2: F0 ω0 = x+ = cos(ω0 t) m xp (t) = t(A cos(ω0 t) + B sin(ω0 t)) ￿￿ ￿ xp (t) 2 ω0 x ￿ k m = A cos(ω0 t) + B sin(ω0 t) +t(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) x￿￿ (t) = −ω0 A sin(ω0 t) + ω0 B cos(ω0 t) p +(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) A=0 F0 F0 B= =√ 2ω 0 m 2 km 2 2 +t(−ω0 A cos(ω0 t) − ω0 B sin(ω0 t)) Forced vibrations (3.8) ω = ω0 • Case 2: F0 ω0 = x+ = cos(ω0 t) m xp (t) = t(A cos(ω0 t) + B sin(ω0 t)) ￿￿ ￿ xp (t) 2 ω0 x ￿ k m = A cos(ω0 t) + B sin(ω0 t) +t(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) x￿￿ (t) = −ω0 A sin(ω0 t) + ω0 B cos(ω0 t) p +(−ω0 A sin(ω0 t) + ω0 B cos(ω0 t)) A=0 F0 F0 B= =√ 2ω 0 m 2 km 2 2 +t(−ω0 A cos(ω0 t) − ω0 B sin(ω0 t)) F0 xp (t) = √ t sin(ω0 t) 2 km Forced vibrations (3.8) • Plot of the amplitude of the particular solution as a function of ω. • Calculated: F0 A= 2 m(ω0 − ω 2 ) • Plotted: • Recall that for ω = ω0 , the amplitude grows without bound. Forced vibrations (3.8) • With damping (on the blackboard) Review questions Review questions Review questions • A dye diffuses between two chambers at a rate proportional to the difference in concentrations (c1 and c2) between the chambers (with proportionality constant k>0). Write down a differential equation for the concentration in the ﬁrst chamber. • Solve: y ￿ − 2ty = t Review questions • A dye diffuses between two chambers at a rate proportional to the difference in concentrations (c1 and c2) between the chambers (with proportionality constant k>0). Write down a differential equation for the concentration in the ﬁrst chamber. dc1 = k (c2 − c1 ) dt • Solve: y ￿ − 2ty = t Review questions • A dye diffuses between two chambers at a rate proportional to the difference in concentrations (c1 and c2) between the chambers (with proportionality constant k>0). Write down a differential equation for the concentration in the ﬁrst chamber. dc1 = k (c2 − c1 ) dt • Solve: y ￿ − 2ty =...
View Full Document

## This note was uploaded on 02/12/2014 for the course MATH 256 taught by Professor Ericcytrynbaum during the Spring '13 term at The University of British Columbia.

Ask a homework question - tutors are online