Linearizing systems of differential equations
Spring Equation with Forcing In a forced harmonic oscillator, an external forcing function acts, changing the motion. ⎡− 1 1 ⎢0 2 ⎣
Net force = spring restoring force + damping force + driving force (mass)(acceleration) = - (spring constant)(displacement) 3⎤ (damping constant)(velocity) + external force − 2⎥⎦ The negative signs indicate the force opposes the motion.
my'' + βy' + ky = f(t) To solve the differential equation for a forcing function acting on a mass-spring system we will use the method of undetermined coefficients. Butler CC Math Friesen
Driven Mass Spring System
Equilibrium point y = 0 (resting position) f(t) stretching is positive displacement and velocity The dependent variable y represents the position of the mass at time t. y' is the velocity at time t. Butler CC Math Friesen
Spring Equation Example 1
⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
Suppose a weight of 32 lb stretches a spring 8 inches. It is acted on by a constant force of 8 lb. The mass is released from rest at a point 4 inches above the equilibrium position. Find the equation of motion and graph it. Weight = (mass)(g) so mass = 32/32 = 1 slug k = force/displacement = 32/8 = 4 lb/in y(0) = -4 in; y'(0) = 0 in/sec
Differential equation of displacement y with initial conditons: y'' + 4y = 8 y(0) = -4; y'(0) = 0 Butler CC Math Friesen
Spring Equation Example 1 Solving:
y'' + 4y = 8 y(0) = -4; y'(0) = 0
Homogeneous roots: s2 + 4 = 0 s = ±2i Nonhomogeneous root: s* = 0 ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
yh = k1cos2t + k2sin2t
yp = A
Substituting yp and its derivatives we get 4A = 8; A = 2 y = k1cos2t + k2sin2t + 2
y(0) = -4; y'(0) = 0
y' = -2k1sin2t + 2k2cos2t -4 = k1 + 2; k1 = -6 0 = 2k2; k2 = 0 Butler CC Math Friesen
y = -6cos2t + 2 y' = 12sin2t
Spring Equation Example 1 stretched
compressed
y = -6cos2t + 2 y' = 12sin2t Butler CC Math Friesen
Forced Spring Equation Example 2 my'' + βy' + ky = f(t) A 2 kg mass-spring system driven for an external force 2cost N. The spring constant is 1 N/m; damping = 0.5 kg/s. Write the equation of motion. ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
2y'' + .5y' + 1y = 2cost
Roots of the characteristic equation: s = -0.125 ± 0.696i yh = e-0.125t(k1cos(.696t)+k2sin(.696t)) s* = ± i so yp = Acost + Bsint
Substituting yp into the diff. equation and separating terms: cost terms: -2A + 1/2B + A = 2; sint terms: -2B - 1/2A + B = 0 A = -1.6; B = 0.8 Butler CC Math Friesen
Forced Spring Equation Example 2 yp = -1.6cost + 0.8sint yh = e-0.125t (k1cos(.696t) + k2sin(.696t)) General solution: y = e-0.125t (k1cos(.696t) + k2sin(.696t)) - 1.6cost + 0.8sint ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
y=0 Suppose the mass is released one meter below the equilibrium position with upward velocity 1 m/s. Initial conditions: y(0) = 1; y'(0) = -1
Butler CC Math Friesen
Forced Spring Equation Example 2 General equation of motion: y = e-0.125t (k1cos (.696t) + k2sin(.696t)) - 1.6cost + 0.8sint Find the derivative (velocity): v = y' = -0.125e-0.125t (k1cos(.696t) + k2sin(.696t)) + 0.696e-0.125t (-k1sin(.696t) + k2cos(.696t) + 1.6sint + 0.8cost ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
Substitute initial conditions y(0) = 1 ; y'(0) = -1; find k1, k2:
1 = k1 – 1.6; k1 = 2.6 -1 = -0.125k1 + 0.696k2 + 0.8; k2 = -2.12 Equations of motion with the given initial conditions: y = e-0.125t (2.6cos(.696t) - 2.12sin(.696t) - 1.6cost + 0.8sint v = y' = -0.125e-0.125t (2.6cos(.696t) – 2.12sin(.696t)) + e-0.125t (-2.6sint – 2.12cost) - 1.6sint + 0.8cost
Butler CC Math Friesen
Spring Equation Example 2
⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
CD applet y = e-0.125t (2.6cos(.696t) – 2.12sin(.696t) - 1.6cost + 0.8sint Butler CC Math Friesen
Spring Equation Investigation Tools Brooks/Cole CD Mass Spring ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
Add-Wesley IDE - Tool Index Damped Forced Vibrations
Butler CC Math Friesen
IMES / Mechanics ETH Center Zurich Switzerland
Net force = spring restoring force + damping force + driving force (mass)(acceleration) = - (spring constant)(displacement) 3⎤ (damping constant)(velocity) + external force − 2⎥⎦ The negative signs indicate the force opposes the motion.
my'' + βy' + ky = f(t) To solve the differential equation for a forcing function acting on a mass-spring system we will use the method of undetermined coefficients. Butler CC Math Friesen
Driven Mass Spring System
Equilibrium point y = 0 (resting position) f(t) stretching is positive displacement and velocity The dependent variable y represents the position of the mass at time t. y' is the velocity at time t. Butler CC Math Friesen
Spring Equation Example 1
⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
Suppose a weight of 32 lb stretches a spring 8 inches. It is acted on by a constant force of 8 lb. The mass is released from rest at a point 4 inches above the equilibrium position. Find the equation of motion and graph it. Weight = (mass)(g) so mass = 32/32 = 1 slug k = force/displacement = 32/8 = 4 lb/in y(0) = -4 in; y'(0) = 0 in/sec
Differential equation of displacement y with initial conditons: y'' + 4y = 8 y(0) = -4; y'(0) = 0 Butler CC Math Friesen
Spring Equation Example 1 Solving:
y'' + 4y = 8 y(0) = -4; y'(0) = 0
Homogeneous roots: s2 + 4 = 0 s = ±2i Nonhomogeneous root: s* = 0 ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
yh = k1cos2t + k2sin2t
yp = A
Substituting yp and its derivatives we get 4A = 8; A = 2 y = k1cos2t + k2sin2t + 2
y(0) = -4; y'(0) = 0
y' = -2k1sin2t + 2k2cos2t -4 = k1 + 2; k1 = -6 0 = 2k2; k2 = 0 Butler CC Math Friesen
y = -6cos2t + 2 y' = 12sin2t
Spring Equation Example 1 stretched
compressed
y = -6cos2t + 2 y' = 12sin2t Butler CC Math Friesen
Forced Spring Equation Example 2 my'' + βy' + ky = f(t) A 2 kg mass-spring system driven for an external force 2cost N. The spring constant is 1 N/m; damping = 0.5 kg/s. Write the equation of motion. ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
2y'' + .5y' + 1y = 2cost
Roots of the characteristic equation: s = -0.125 ± 0.696i yh = e-0.125t(k1cos(.696t)+k2sin(.696t)) s* = ± i so yp = Acost + Bsint
Substituting yp into the diff. equation and separating terms: cost terms: -2A + 1/2B + A = 2; sint terms: -2B - 1/2A + B = 0 A = -1.6; B = 0.8 Butler CC Math Friesen
Forced Spring Equation Example 2 yp = -1.6cost + 0.8sint yh = e-0.125t (k1cos(.696t) + k2sin(.696t)) General solution: y = e-0.125t (k1cos(.696t) + k2sin(.696t)) - 1.6cost + 0.8sint ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
y=0 Suppose the mass is released one meter below the equilibrium position with upward velocity 1 m/s. Initial conditions: y(0) = 1; y'(0) = -1
Butler CC Math Friesen
Forced Spring Equation Example 2 General equation of motion: y = e-0.125t (k1cos (.696t) + k2sin(.696t)) - 1.6cost + 0.8sint Find the derivative (velocity): v = y' = -0.125e-0.125t (k1cos(.696t) + k2sin(.696t)) + 0.696e-0.125t (-k1sin(.696t) + k2cos(.696t) + 1.6sint + 0.8cost ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
Substitute initial conditions y(0) = 1 ; y'(0) = -1; find k1, k2:
1 = k1 – 1.6; k1 = 2.6 -1 = -0.125k1 + 0.696k2 + 0.8; k2 = -2.12 Equations of motion with the given initial conditions: y = e-0.125t (2.6cos(.696t) - 2.12sin(.696t) - 1.6cost + 0.8sint v = y' = -0.125e-0.125t (2.6cos(.696t) – 2.12sin(.696t)) + e-0.125t (-2.6sint – 2.12cost) - 1.6sint + 0.8cost
Butler CC Math Friesen
Spring Equation Example 2
⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
CD applet y = e-0.125t (2.6cos(.696t) – 2.12sin(.696t) - 1.6cost + 0.8sint Butler CC Math Friesen
Spring Equation Investigation Tools Brooks/Cole CD Mass Spring ⎡− 1 1 3 ⎤ ⎢ 0 2 − 2⎥ ⎣ ⎦
Add-Wesley IDE - Tool Index Damped Forced Vibrations
Butler CC Math Friesen
IMES / Mechanics ETH Center Zurich Switzerland
