Free mechanical vibrations / Couple mass-spring system

Section 4.9; Section 5.6 Free Mechanical Vibrations/Couple Mass-Spring System

June 30, 2009

Free Mechanical Vibrations/Couple Mass-Spring System

Today’s Session


Today’s Session

A Summary of This Session:


Today’s Session

A Summary of This Session: (1) Free Mechanical Vibration (no forcing term).


Today’s Session

A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems


Today’s Session

A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations


Today’s Session

A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations (4) Eigenvalues and Eigenvectors


Today’s Session

A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations (4) Eigenvalues and Eigenvectors


Free Mechanical Vibrations

Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my ′′ + by ′ + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b.



Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my ′′ + by ′ + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b. Four cases: (1) undamped free case; (2) underdamped case; (3) overdamped case; (4) critical damped case.



Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my ′′ + by ′ + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b. Four cases: (1) undamped free case; (2) underdamped case; (3) overdamped case; (4) critical damped case.


Case 1: Undamped mass-spring system: b = 0. The equation is given by: my ′′ + ky = 0


Case 1: Undamped mass-spring system: b = 0. The equation is given by: my ′′ + ky = 0 or y ′′ +

k y =0 m


Case 1: Undamped mass-spring system: b = 0. The equation is given by: my ′′ + ky = 0 or

k y =0 m q k k 2 Let ω = m . The quantity ω = m is called the angular frequency (measured in radians per second) y ′′ +




Period: T =

2π ω

(measured in seconds)




Period: T = 2π ω (measured in seconds) ω Frequency: f = T1 = 2π (measured in Hertz=1/seconds=# cycles per second)




Period: T = 2π ω (measured in seconds) ω Frequency: f = T1 = 2π (measured in Hertz=1/seconds=# cycles per second) The solution is given by: y = C1 cos ωt + C2 sin ωt = A sin(ωt + φ).


Undamped case, cont’d Here A is called the amplitude, and φ is called the phase. They are given by q C1 A = C12 + C22 φ = arctan . C2


Undamped case, cont’d Here A is called the amplitude, and φ is called the phase. They are given by q C1 A = C12 + C22 φ = arctan . C2 Example A 1/8 kg manss is attached to a spring with stiffness k = 16N/m. The mass is displaed 0.5 m to the right of the √ equilibrium point and given an outward initial velocity of 2 m/s. (a) Neglecting damping, find a formula of the displacement as a function of time. Display the values of ω, T , f , A and φ.


Undamped case, cont’d Here A is called the amplitude, and φ is called the phase. They are given by q C1 A = C12 + C22 φ = arctan . C2 Example A 1/8 kg manss is attached to a spring with stiffness k = 16N/m. The mass is displaed 0.5 m to the right of the √ equilibrium point and given an outward initial velocity of 2 m/s. (a) Neglecting damping, find a formula of the displacement as a function of time. Display the values of ω, T , f , A and φ. (b) How long after release does the mass pass first through the equilibrium position?


Undamped case, cont’d Answers: √ (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec.


Undamped case, cont’d Answers: √ (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec. y (t) =

√ √ 1 1 cos 8 2t + sin 8 2t 2 8


Undamped case, cont’d Answers: √ (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec. y (t) = A=

q

√ √ 1 1 cos 8 2t + sin 8 2t 2 8

√

1/2 and tan φ = 1/8 = 4 so φ = 1.326 rad. √ √ 17 sin(8 2t + 1.326) y (t) = 8

( 12 )2 + ( 81 )2 =

17 8


Undamped case, cont’d Answers: √ (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec. y (t) = A=

q

√ √ 1 1 cos 8 2t + sin 8 2t 2 8

√

1/2 and tan φ = 1/8 = 4 so φ = 1.326 rad. √ √ 17 sin(8 2t + 1.326) y (t) = 8

( 12 )2 + ( 81 )2 =

17 8

√ √ (b) y (t) = 0 means sin(8 2t + φ) = 0 so 8 2t + φ = kπ for k = 0.16 sec. Every 1/2 period (or integer. Solving gives: t = kπ−φ ω 0.28 sec) the mass goes through the equilibrium point.


Overdamped case: b 2 − 4k m > 0 The equation is my ′′ + by ′ + ky = 0


Overdamped case: b 2 − 4k m > 0 The equation is my ′′ + by ′ + ky = 0 The characteristic equation is given by mr 2 + br + k = 0


Overdamped case: b 2 − 4k m > 0 The equation is my ′′ + by ′ + ky = 0 The characteristic equation is given by mr 2 + br + k = 0 so b ± r =− 2m The solution is given by:

√

b 2 − 4k m 2m

y (t) = C1 e r1 t + C2 e r2 t


Overdamped case: b 2 − 4k m > 0 The equation is my ′′ + by ′ + ky = 0 The characteristic equation is given by mr 2 + br + k = 0 so b ± r =− 2m The solution is given by:

√

b 2 − 4k m 2m

y (t) = C1 e r1 t + C2 e r2 t Note that both r1 and r2 are negative (why?), so as t → ∞, y (t) → 0. Free Mechanical Vibrations/Couple Mass-Spring System

Underdamped case: b 2 − 4k m < 0 With ω =

√

4k m−b2 , 2m

the roots of the characteristic equation are: r =−

b ±iω 2m

So the solution is given by: b

b

y (t) = C1 e − 2m t cos ωt + C2 e − 2m t sin ωt.



√

4k m−b2 , 2m


b ±iω 2m


b

y (t) = C1 e − 2m t cos ωt + C2 e − 2m t sin ωt. Over time y (t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency.



√

4k m−b2 , 2m


b ±iω 2m


b

y (t) = C1 e − 2m t cos ωt + C2 e − 2m t sin ωt. Over time y (t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency. The solution can be put in the form b

y (t) = A e − 2m t sin(ωt + φ) where (as before): A=

q

C12 + C22

φ = arctan

C1 . C2



√

4k m−b2 , 2m


b ±iω 2m


b

y (t) = C1 e − 2m t cos ωt + C2 e − 2m t sin ωt. Over time y (t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency. The solution can be put in the form b

y (t) = A e − 2m t sin(ωt + φ) where (as before): A=

q

C12 + C22

φ = arctan

C1 . C2


Critically damped case: b 2 − 4k m = 0 In this case, the characteristic equation has a double root b , So the solution is r = − 2m b

b

y (t) = C1 e − 2m t + C2 t e − 2m t



b

y (t) = C1 e − 2m t + C2 t e − 2m t Example: (related to webassign question and example p. 234)



b

y (t) = C1 e − 2m t + C2 t e − 2m t Example: (related to webassign question and example p. 234) Find the value of b for which y ′′ + b y ′ + 25y = 0 y (0) = 1,

y ′ (0) = 0

is critically damped. Solve for y (t) in this case and sketch it.



b

y (t) = C1 e − 2m t + C2 t e − 2m t Example: (related to webassign question and example p. 234) Find the value of b for which y ′′ + b y ′ + 25y = 0 y (0) = 1,

y ′ (0) = 0

is critically damped. Solve for y (t) in this case and sketch it. Answer: b = 10; y (t) = (1 − t)e −5t


5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p. 308.


5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p. 308. I will show the derivation of the following equations in class:


5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p. 308. I will show the derivation of the following equations in class: We wish to solve: n m x ′′ = −k x + k (y − x) 1 1 2 m2 y ′′ = −k2 (y − x)


5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p. 308. I will show the derivation of the following equations in class: We wish to solve: n m x ′′ = −k x + k (y − x) 1 1 2 m2 y ′′ = −k2 (y − x) In our example: m1 = 2kg ; k1 = 4N/m; m2 = 1kg ; and k2 = 2N/m. So: n 2x ′′ = −6x + 2y y ′′ = 2x − 2y


5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p. 308. I will show the derivation of the following equations in class: We wish to solve: n m x ′′ = −k x + k (y − x) 1 1 2 m2 y ′′ = −k2 (y − x) In our example: m1 = 2kg ; k1 = 4N/m; m2 = 1kg ; and k2 = 2N/m. So: n 2x ′′ = −6x + 2y y ′′ = 2x − 2y


Example 1, p. 308, cont’d Let’s clean this up a little bit: n x ′′ = −3x + y y ′′ = 2x − 2y


Example 1, p. 308, cont’d Let’s clean this up a little bit: n x ′′ = −3x + y y ′′ = 2x − 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: n (D 2 + 3)x − y = 0 −2x + (D 2 + 2)y

= 0


Example 1, p. 308, cont’d Let’s clean this up a little bit: n x ′′ = −3x + y y ′′ = 2x − 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: n (D 2 + 3)x − y = 0 Therefore

−2x + (D 2 + 2)y

= 0

(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0



−2x + (D 2 + 2)y

= 0

(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0

We note that: (D 2 + 2)y = 2x.



−2x + (D 2 + 2)y

= 0

(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0

We note that: (D 2 + 2)y = 2x. So: (D 2 + 2)(D 2 + 3)x − 2x = 0.



−2x + (D 2 + 2)y

= 0

(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0


We clean this up a little bit:

(D 4 + 5D 2 + 4)x = 0. Free Mechanical Vibrations/Couple Mass-Spring System


−2x + (D 2 + 2)y

= 0

(D 2 + 2)(D 2 + 3)x − (D 2 + 2)y = 0


We clean this up a little bit:

(D 4 + 5D 2 + 4)x = 0. Free Mechanical Vibrations/Couple Mass-Spring System

Example 1, p. 308, cont’d

Then factor: (D 2 + 1)(D 2 + 4)x = 0


Example 1, p. 308, cont’d

Then factor: (D 2 + 1)(D 2 + 4)x = 0 Therefore: x(t) = C1 cos t + C2 sin t + C3 cos 2t + C4 sin 2t and y = (D 2 + 3)x = 2C1 cos t + 2C2 sin t − C3 cos 2t − C4 sin 2t


Example 2

Use the method of the previous example (the annihilator method) to solve the first order (homogeneous) linear system: n x ′ = 3x + 4y y ′ = 4x + 3y


Example 2

Use the method of the previous example (the annihilator method) to solve the first order (homogeneous) linear system: n x ′ = 3x + 4y y ′ = 4x + 3y Answer: x(t) = C1 e 7t + C2 e −t and y (t) = C1 e −t − C2 e −t
