Differential Equations Second-Order Linear DEs Mechanical Oscillations Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A couple of physical situations can be modeled by 2nd order D. E A couple of physical situations can be modeled by 2nd order D.E.s with constant coefficients. Mechanical oscillations, such as a mass-spring setup, and an analogous electrical system involving resistors, capacitors and inductors. Consider the mass-spring system shown. The coordinates are defined so that the equilibrium position is y=0, and down is positive. Using Newton’s 2nd law of motion we obtain an equation involving the forces on the mass: m=mass g=damping coefficient k=spring constant f(t)=external driving force Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Case 1: Δ<0 Underdamped Here is the basic form of a homogeneous 2nd order linear O.D.E. that models the motion of a mass/spring setup. The solution can take 3 different forms, depending on the roots of the characteristic equation. The inside of the square root is called the discriminant (symbol is Δ) Case 1: Δ<0 Underdamped In this case the roots are complex, and the solutions oscillate about the equilibrium and decay exponentially due to damping.

Case 2: Δ=0 Critically Damped Here is the basic form of a homogeneous 2nd order linear O.D.E. that models the motion of a mass/spring setup. The solution can take 3 different forms, depending on the roots of the characteristic equation. The inside of the square root is called the discriminant (symbol is Δ) Case 2: Δ=0 Critically Damped In this case the root is repeated, and the solutions pass through the equilibrium at most once.

Case 3: Δ>0 OverDamped Here is the basic form of a homogeneous 2nd order linear O.D.E. that models the motion of a mass/spring setup. The solution can take 3 different forms, depending on the roots of the characteristic equation. The inside of the square root is called the discriminant (symbol is Δ) Case 3: Δ>0 OverDamped In this case there are 2 distinct real roots. If both roots are negative, the solution passes through equilibrium at most once.

Example 1: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=sin(2t) is applied and the mass starts at the equilibrium with velocity 2 m/s. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=sin(2t) is applied and the mass starts at the equilibrium with velocity 2 m/s. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

First the homogeneous solution: Example 1: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=sin(2t) is applied and the mass starts at the equilibrium with velocity 2 m/s. First the homogeneous solution: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

First the homogeneous solution: Example 1: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=sin(2t) is applied and the mass starts at the equilibrium with velocity 2 m/s. First the homogeneous solution: Next the particular solution. Using undetermined coefficients: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

First the homogeneous solution: Example 1: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=sin(2t) is applied and the mass starts at the equilibrium with velocity 2 m/s. First the homogeneous solution: Next the particular solution. Using undetermined coefficients: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Finally plug in the initial values: Example 1: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=sin(2t) is applied and the mass starts at the equilibrium with velocity 2 m/s. Finally plug in the initial values: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Finally plug in the initial values: Example 1: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=sin(2t) is applied and the mass starts at the equilibrium with velocity 2 m/s. Finally plug in the initial values: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 2 - Beats: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=cos(.9t) is applied and the mass begins at rest at the equilibrium position. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 2 - Beats: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=cos(.9t) is applied and the mass begins at rest at the equilibrium position. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 2 - Beats: A 1kg mass is attached to a spring with constant k=1 N/m and there is negligible damping. An external driving force f(t)=cos(.9t) is applied and the mass begins at rest at the equilibrium position. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 3: A mass of 5kg stretches a spring 10cm Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 3: A mass of 5kg stretches a spring 10cm Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 3: A mass of 5kg stretches a spring 10cm Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 3: A mass of 5kg stretches a spring 10cm Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. Divide by 5 to make numbers easier. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 3: A mass of 5kg stretches a spring 10cm Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. Starts at equilibrium position Initial velocity (convert to m/s) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

First solve the homogeneous equation: Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. First solve the homogeneous equation: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

First solve the homogeneous equation: Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. First solve the homogeneous equation: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Next find the particular solution – using undetermined coefficients: Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. Next find the particular solution – using undetermined coefficients: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Next find the particular solution – using undetermined coefficients: Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. Next find the particular solution – using undetermined coefficients: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Finally we can use the initial values to find the c1 and c2. Example 3: A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of 10sin(t/2)Newtons and moves in a medium that imparts a viscous force of 2N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate and solve the initial value problem describing the motion of the mass. Finally we can use the initial values to find the c1 and c2. Our final answer: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 4: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this in the usual way with the characteristic equation. Example 4: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. For this unit system we will take the acceleration due to gravity to be 32 ft/s2. To find the MASS of the object we need to divide the weight by 32. Thus we have m=1. Solve this in the usual way with the characteristic equation. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this in the usual way with the characteristic equation. Example 4: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. For this unit system we will take the acceleration due to gravity to be 32 ft/s2. To find the MASS of the object we need to divide the weight by 32. Thus we have m=1. Solve this in the usual way with the characteristic equation. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this in the usual way with the characteristic equation. Example 4: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. For this unit system we will take the acceleration due to gravity to be 32 ft/s2. To find the MASS of the object we need to divide the weight by 32. Thus we have m=1. Solve this in the usual way with the characteristic equation. Plug in initial values: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 4: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. The solution is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 4: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. The solution is Using some trigonometry this can be re-written as a single cosine function. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 4: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. The solution is Using some trigonometry this can be re-written as a single cosine function. The plot of the solution shows the function oscillating between the exponential “envelope” and eventually approaching 0. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 5: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The oscillation is driven with an external force given by the function F(t)=2sin(3t). The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 5: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The oscillation is driven with an external force given by the function F(t)=2sin(3t). The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. For this unit system we will take the acceleration due to gravity to be 32 ft/s2. To find the MASS of the object we need to divide the weight by 32. Thus we have m=1. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the homogeneous equation first: Example 5: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The oscillation is driven with an external force given by the function F(t)=2sin(3t). The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. For this unit system we will take the acceleration due to gravity to be 32 ft/s2. To find the MASS of the object we need to divide the weight by 32. Thus we have m=1. Solve the homogeneous equation first: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the homogeneous equation first: Example 5: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The oscillation is driven with an external force given by the function F(t)=2sin(3t). The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. For this unit system we will take the acceleration due to gravity to be 32 ft/s2. To find the MASS of the object we need to divide the weight by 32. Thus we have m=1. Solve the homogeneous equation first: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the homogeneous equation first: Example 5: An object weighing 32 pounds is hung from a spring with constant 13 lb/ft. There is a damping mechanism with resistance coefficient 4 lb-s/ft. The oscillation is driven with an external force given by the function F(t)=2sin(3t). The object is initially 2 ft. above the equilibrium position and its velocity is 1 ft/s downwards. Formulate the O.D.E. for this situation and solve for the displacement function. For this unit system we will take the acceleration due to gravity to be 32 ft/s2. To find the MASS of the object we need to divide the weight by 32. Thus we have m=1. Solve the homogeneous equation first: The homogeneous solution is: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

The homogeneous solution is: A guess for the particular solution is Compare to the homogeneous solution and none of the terms are repeated here. The exponential term makes the homogeneous solutions linear independent from these. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

The homogeneous solution is: A guess for the particular solution is Compare to the homogeneous solution and none of the terms are repeated here. The exponential term makes the homogeneous solutions linear independent from these. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

The homogeneous solution is: A guess for the particular solution is Compare to the homogeneous solution and none of the terms are repeated here. The exponential term makes the homogeneous solutions linear independent from these. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

The homogeneous solution is: A guess for the particular solution is Compare to the homogeneous solution and none of the terms are repeated here. The exponential term makes the homogeneous solutions linear independent from these. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

The homogeneous solution is: A guess for the particular solution is Compare to the homogeneous solution and none of the terms are repeated here. The exponential term makes the homogeneous solutions linear independent from these. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

The general solution is: Plug in the given values to find c1 and c2. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Steady-State solution Transient solution Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Some useful trig identities for this type of problem: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Electric Circuits Another physical system with the same oscillatory nature as the mass/spring setup is the circuit pictured below. A resistor, capacitor and inductor are wired in series, with a voltage driving current around the circuit. Using Kirchoff’s Laws we can obtain the following DE: This is written in terms of the charge on the capacitor, Q(t). Alternatively, we could write this in terms of the current, I(t): Note that these equations have the same form as the mass/spring setups that we saw previously, so the solution methods are the same. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB