Spring Eqn w/ Friction Consider the equation: m d 2 y dt 2 + b dy dt - - PowerPoint PPT Presentation

Spring Eqn w/ Friction

Consider the equation: m d2y

dt2 + b dy dt + ky = 0

A spring w/ stiffness k, friction b and mass m, all positive values

Free Vibrations – p. 1/5

Spring Eqn w/ Friction

Consider the equation: m d2y

dt2 + b dy dt + ky = 0

A spring w/ stiffness k, friction b and mass m, all positive values The characteristic equation is: mr2 + br + y = 0 which has roots r = −b±

√ b2−4mk 2m

= − b

2m ± 1 2m

√ b2 − 4mk

Free Vibrations – p. 1/5

Spring Eqn w/ Friction

Consider the equation: m d2y

dt2 + b dy dt + ky = 0

A spring w/ stiffness k, friction b and mass m, all positive values The characteristic equation is: mr2 + br + y = 0 which has roots r = −b±

√ b2−4mk 2m

= − b

2m ± 1 2m

√ b2 − 4mk So yH(t) = c1e− b

2m + 1 2m

√ b2−4mkt + c2e− b

2m − 1 2m

√ b2−4mkt

Free Vibrations – p. 1/5

Underdamped Vibrations

If the friction coefficient, b, is small enough so that the discriminant b2 − 4mk < 0, the result is underdamped vibrations - a sinusoidal vibration whose size is exponentially dying.

Free Vibrations – p. 2/5

Underdamped Vibrations

If the friction coefficient, b, is small enough so that the discriminant b2 − 4mk < 0, the result is underdamped vibrations - a sinusoidal vibration whose size is exponentially dying. The roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {− b

2m + i 2m

√ 4mk − b2, − b

2m − i 2m

√ 4mk − b2}

Free Vibrations – p. 2/5

Underdamped Vibrations

If the friction coefficient, b, is small enough so that the discriminant b2 − 4mk < 0, the result is underdamped vibrations - a sinusoidal vibration whose size is exponentially dying. The roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {− b

2m + i 2m

√ 4mk − b2, − b

2m − i 2m

√ 4mk − b2} So yH(t) = c1e− b

2m + 1 2m

√ b2−4mkt + c2e− b

2m − 1 2m

√ b2−4mkt

Free Vibrations – p. 2/5

Rewriting as sin/cos and phase

Recall Euler’s Equation: eit = cos(t) + i sin(t). If the roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {a + bi, a − bi} Then c1ea+bi + c2ea−bi can be rewritten as:

Free Vibrations – p. 3/5

Rewriting as sin/cos and phase

Recall Euler’s Equation: eit = cos(t) + i sin(t). If the roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {a + bi, a − bi} Then c1ea+bi + c2ea−bi can be rewritten as: c1(eaebi) + c2(eae−bi)

Free Vibrations – p. 3/5

Rewriting as sin/cos and phase

Recall Euler’s Equation: eit = cos(t) + i sin(t). If the roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {a + bi, a − bi} Then c1ea+bi + c2ea−bi can be rewritten as: c1(eaebi) + c2(eae−bi) = ea(c1ebi + c2e−bi)

Free Vibrations – p. 3/5

Rewriting as sin/cos and phase

Recall Euler’s Equation: eit = cos(t) + i sin(t). If the roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {a + bi, a − bi} Then c1ea+bi + c2ea−bi can be rewritten as: c1(eaebi) + c2(eae−bi) = ea(c1ebi + c2e−bi) = ea(c1(cos(b) + i sin(b)) + c2(cos(−b) + i sin(−b)))

Free Vibrations – p. 3/5

Rewriting as sin/cos and phase

Recall Euler’s Equation: eit = cos(t) + i sin(t). If the roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {a + bi, a − bi} Then c1ea+bi + c2ea−bi can be rewritten as: c1(eaebi) + c2(eae−bi) = ea(c1ebi + c2e−bi) = ea(c1(cos(b) + i sin(b)) + c2(cos(−b) + i sin(−b))) = ea(c1(cos(b) + i sin(b)) + c2(cos(b) − i sin(b)))

Free Vibrations – p. 3/5

Rewriting as sin/cos and phase

Recall Euler’s Equation: eit = cos(t) + i sin(t). If the roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {a + bi, a − bi} Then c1ea+bi + c2ea−bi can be rewritten as: c1(eaebi) + c2(eae−bi) = ea(c1ebi + c2e−bi) = ea(c1(cos(b) + i sin(b)) + c2(cos(−b) + i sin(−b))) = ea(c1(cos(b) + i sin(b)) + c2(cos(b) − i sin(b))) = ea((c1 + c2) cos(b) + i(c1 − c2) sin(b)) = ea(d1 cos(b) + d2 sin(b)) (where d1 = c1 + c2 and d2 = i(c1 − c2))

Free Vibrations – p. 3/5

Rewriting as sin/cos and phase

Recall Euler’s Equation: eit = cos(t) + i sin(t). If the roots of the characteristic polynomial are a complex conjugate pair: {r1, r2} = {a + bi, a − bi} Then c1ea+bi + c2ea−bi can be rewritten as: c1(eaebi) + c2(eae−bi) = ea(c1ebi + c2e−bi) = ea(c1(cos(b) + i sin(b)) + c2(cos(−b) + i sin(−b))) = ea(c1(cos(b) + i sin(b)) + c2(cos(b) − i sin(b))) = ea((c1 + c2) cos(b) + i(c1 − c2) sin(b)) = ea(d1 cos(b) + d2 sin(b)) (where d1 = c1 + c2 and d2 = i(c1 − c2)) Also, = Aea cos(b + φ) (where A =

• d2

1 + d2 2 and φ = arctan(d1 d2))

Free Vibrations – p. 3/5

Underdamped (cont)

If the friction coefficient, b, is small enough so that the discriminant b2 − 4mk < 0, yH(t) = c1e(−

b 2m + i 2m

√ 4mk−b2)t + c2e(−

b 2m − i 2m

√ 4mk−b2)t

Free Vibrations – p. 4/5

Underdamped (cont)

If the friction coefficient, b, is small enough so that the discriminant b2 − 4mk < 0, yH(t) = c1e(−

b 2m + i 2m

√ 4mk−b2)t + c2e(−

b 2m − i 2m

√ 4mk−b2)t

= e−

b 2m t(c1e it 2m

√ 4mk−b2 + c2e− it

2m

√ 4mk−b2)

Free Vibrations – p. 4/5

Underdamped (cont)

If the friction coefficient, b, is small enough so that the discriminant b2 − 4mk < 0, yH(t) = c1e(−

b 2m + i 2m

√ 4mk−b2)t + c2e(−

b 2m − i 2m

√ 4mk−b2)t

= e−

b 2m t(c1e it 2m

√ 4mk−b2 + c2e− it

2m

√ 4mk−b2)

= e−

b 2m t((c1 + c2) cos( t

2m

√ 4mk − b2) + i(c1 − c2) sin( t

2m

√ 4mk − b2)) = e−

b 2m t(d1 cos( t

2m

√ 4mk − b2) + d2 sin( t

2m

√ 4mk − b2))

Free Vibrations – p. 4/5

Underdamped (cont)

If the friction coefficient, b, is small enough so that the discriminant b2 − 4mk < 0, yH(t) = c1e(−

b 2m + i 2m

√ 4mk−b2)t + c2e(−

b 2m − i 2m

√ 4mk−b2)t

= e−

b 2m t(c1e it 2m

√ 4mk−b2 + c2e− it

2m

√ 4mk−b2)

= e−

b 2m t((c1 + c2) cos( t

2m

√ 4mk − b2) + i(c1 − c2) sin( t

2m

√ 4mk − b2)) = e−

b 2m t(d1 cos( t

2m

√ 4mk − b2) + d2 sin( t

2m

√ 4mk − b2)) = Ae−

b 2m t sin( t

2m

√ 4mk − b2 + φ)

Free Vibrations – p. 4/5

Underdamped (cont)

If the friction coefficient, b, is small enough so that the discriminant b2 − 4mk < 0, yH(t) = c1e(−

b 2m + i 2m

√ 4mk−b2)t + c2e(−

b 2m − i 2m

√ 4mk−b2)t

= e−

b 2m t(c1e it 2m

√ 4mk−b2 + c2e− it

2m

√ 4mk−b2)

= e−

b 2m t((c1 + c2) cos( t

2m

√ 4mk − b2) + i(c1 − c2) sin( t

2m

√ 4mk − b2)) = e−

b 2m t(d1 cos( t

2m

√ 4mk − b2) + d2 sin( t

2m

√ 4mk − b2)) = Ae−

b 2m t sin( t

2m

√ 4mk − b2 + φ)

• 10
• 5

5 10

• 2
• 1

1 2

Free Vibrations – p. 4/5

Overdamped Case

If the friction coefficient, b, is large enough so that the discriminant b2 − 4mk > 0, the result is overdamped - an exponentially dying curve crossing the axis at most once.

Free Vibrations – p. 5/5

Overdamped Case

If the friction coefficient, b, is large enough so that the discriminant b2 − 4mk > 0, the result is overdamped - an exponentially dying curve crossing the axis at most once.

The roots of the characteristic polynomial are two real numbers, both negative: {r1, r2} = {− b

2m + i 2m

√ 4mk − b2, − b

2m − i 2m

√ 4mk − b2}

Free Vibrations – p. 5/5

Overdamped Case

If the friction coefficient, b, is large enough so that the discriminant b2 − 4mk > 0, the result is overdamped - an exponentially dying curve crossing the axis at most once.

The roots of the characteristic polynomial are two real numbers, both negative: {r1, r2} = {− b

2m + i 2m

√ 4mk − b2, − b

2m − i 2m

√ 4mk − b2} So yH(t) = c1e−

b 2m + 1 2m

√ b2−4mkt + c2e−

b 2m − 1 2m

√ b2−4mkt

= c1er1t + c2er2t

Free Vibrations – p. 5/5

Overdamped Case

If the friction coefficient, b, is large enough so that the discriminant b2 − 4mk > 0, the result is overdamped - an exponentially dying curve crossing the axis at most once.

The roots of the characteristic polynomial are two real numbers, both negative: {r1, r2} = {− b

2m + i 2m

√ 4mk − b2, − b

2m − i 2m

√ 4mk − b2} So yH(t) = c1e−

b 2m + 1 2m

√ b2−4mkt + c2e−

b 2m − 1 2m

√ b2−4mkt

= c1er1t + c2er2t

2 4 6 8

• 0.2
• 0.1

0.1 0.2 0.3 0.4

Free Vibrations – p. 5/5