# sln4 - 7224 Homework Solution 4 Problem 1(a m C sin = A sin...

This preview shows pages 1–5. Sign up to view the full content.

7224 Homework Solution 4 Problem 1 (a) m ¨ θ + α ˙ θ + C sin θ = A sin ω 0 t Let t = c t ¯ t , the equation changes into d 2 θ d ¯ t 2 + α m c t d ¯ t + C c 2 t m sinθ = A m c 2 t sin ( ω 0 c t ¯ t ) (b) To eliminate a second coeﬃcient, we have three possible ways i) Let αc t /m = 1 and c t = m/α , which is the time scale for exponential decay. The equation changes into d 2 θ d ¯ t 2 + d ¯ t + C m α 2 sin θ = Am α 2 sin( ω 0 m α ¯ t ) ii) Let Cc 2 t /m = 1 and c t = q m/C , which is the time scale for the oscillation.The equation changes into d 2 θ d ¯ t 2 + α Cm d ¯ t + sin θ = A C sin( ω 0 r m C ¯ t ) iii) Let Ac 2 t /m = 1 and c t = q m/A , which is the time scale for the driving.The equation changes into d 2 θ d ¯ t 2 + α Am d ¯ t + C A sin θ = sin( ω 0 r m A ¯ t ) (c) From the equations above, Am α 2 is the Rayleigh-like parameter with the meaning of “ratio of strength of drinving to strength of dissipation.” and c t = m/α is the only choice. (d) For a damped pendulum system X 00 + βX 0 + ω 2 X = 0 The general solution is approximately X ( t ) C 0 e - βt/ 2 sinωt 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
and the quality factor can be deﬁne as Q = ω β where in this system

This preview has intentionally blurred sections. Sign up to view the full version.

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

{[ snackBarMessage ]}

### Page1 / 5

sln4 - 7224 Homework Solution 4 Problem 1(a m C sin = A sin...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online