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20.8.7 Again solve Eq. (20.160), which describes a damped simple harmonic oscillator, but this time for X (0) = 0, X'(0) = vo, and (a) b² < 4mk (underdamped), (b) b²4mk (critically damped), (c) b²> 4mk (overdamped). VO w1 (b) X(t)= vote ANS. (a) X(t) = -e-(b/2m) sint, -(b/2m)1

20.8.7 Again solve Eq. (20.160), which describes a damped simple harmonic oscillator, but this time for X (0) = 0, X'(0) = vo, and (a) b² < 4mk (underdamped), (b) b²4mk (critically damped), (c) b²> 4mk (overdamped). VO w1 (b) X(t)= vote ANS. (a) X(t) = -e-(b/2m) sint, -(b/2m)1

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Example 20.8.5 DAMPED OSCILLATOR Equations (20.158) and (20.159) are useful when we consider an oscillating mass with damping proportional to the velocity. Equation (20.151), with such damping added, becomes mX" (1) +bX' (1) +kX (1) = 0, (20.160) in which b is a proportionality constant. Let us assume that the particle starts from rest at X (0) = Xo, so X'(0) = 0. The transformed equation is m[s²x(s)-sXo] + b[sx(s) - Xo]+kx(s) = 0, with solution x(s)= Xo- This transform does not appear in our table, but may be handled by completing the square of the denominator: m ms+b ms²+bs+k 2 k b + 2 = ( ² + 2 ) ² + ( = -5²). m 2m m 4m² Considering further only the case that the damping is small enough that b² <4km, then the last term is positive and will be denoted by . We then rearrange x(s) to the form x(s) = Xo s+b/m (s+b/2m)²+w s+b/2m (s+b/2m)² + =Xo (b/2mw) (s+b/2m)² + w These are the same transforms we encountered in Eqs. (20.158) and (20.159), so we may take the inverse transform of our formula for x(s), reaching b X(t)= Xoe-(b/2m) (cosant + sin 2ma Here we have made the substitutions 000 = Xo- e-b/2m) cos(at-9). tano = +Xo b 2ma w = 201 (20.161) Of course, as b→ 0, this solution goes over to the undamped solution, given in Example 20.8.2.
Transcribed Image Text:Example 20.8.5 DAMPED OSCILLATOR Equations (20.158) and (20.159) are useful when we consider an oscillating mass with damping proportional to the velocity. Equation (20.151), with such damping added, becomes mX" (1) +bX' (1) +kX (1) = 0, (20.160) in which b is a proportionality constant. Let us assume that the particle starts from rest at X (0) = Xo, so X'(0) = 0. The transformed equation is m[s²x(s)-sXo] + b[sx(s) - Xo]+kx(s) = 0, with solution x(s)= Xo- This transform does not appear in our table, but may be handled by completing the square of the denominator: m ms+b ms²+bs+k 2 k b + 2 = ( ² + 2 ) ² + ( = -5²). m 2m m 4m² Considering further only the case that the damping is small enough that b² <4km, then the last term is positive and will be denoted by . We then rearrange x(s) to the form x(s) = Xo s+b/m (s+b/2m)²+w s+b/2m (s+b/2m)² + =Xo (b/2mw) (s+b/2m)² + w These are the same transforms we encountered in Eqs. (20.158) and (20.159), so we may take the inverse transform of our formula for x(s), reaching b X(t)= Xoe-(b/2m) (cosant + sin 2ma Here we have made the substitutions 000 = Xo- e-b/2m) cos(at-9). tano = +Xo b 2ma w = 201 (20.161) Of course, as b→ 0, this solution goes over to the undamped solution, given in Example 20.8.2.
20.8.7 Again solve Eq. (20.160), which describes a damped simple harmonic oscillator, but this time for X (0) = 0, X'(0) = vo, and (a) b² < 4mk (underdamped), (b) b²4mk (critically damped), (c) b²> 4mk (overdamped). ANS. (a) X(t) = (b) X(t)= VO w1 vote e-(b/2m) sincit, -(b/2m)
Transcribed Image Text:20.8.7 Again solve Eq. (20.160), which describes a damped simple harmonic oscillator, but this time for X (0) = 0, X'(0) = vo, and (a) b² < 4mk (underdamped), (b) b²4mk (critically damped), (c) b²> 4mk (overdamped). ANS. (a) X(t) = (b) X(t)= VO w1 vote e-(b/2m) sincit, -(b/2m)
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