0:19

We pushed the mass up and release it from the rest

at a point 50 cm above the equilibrium position, okay?

Assume that the gravitational acceleration is g = 10 m per second squared, okay?

And that the damping constant of the medium

surrounding the mass is 7 kg per second.

And under this situation, we'd like to find

the displacement x(t) of the mass, okay?

And we would like to see that whether the mass can pass through the equilibrium

position or not, okay?

1:10

Since we have m is equal to 1 kilogram.

And we take the g is equal 10, okay?

So mg = 10 kilograms per meter per second squared.

And that must be equal to k times elongation, right?

1:32

Initially, the spring's length is 5, and

when the mass is are attached, it's length is equal to 6.

So the elongation is 6 minus 5, that is equal to 1 meter, right?

So from this equation, you can get stiffness of the spring

constant k = 10 kilograms per second square, okay?

So that the displacement, x(t),

must satisfy the initial value problem, okay?

2:08

First, the differential equation x double

prime + 7x prime, okay?

7 is the damping constant given and 10 times of x = 0.

10 is the spring constant, right?

And initially it is stretched further upward 50 cm,

so that x(0) will be- 0.5, and

because it's released from the rest,

the initial velocity is equal to 0, okay?

So that's the initial value problem governing the situation, okay?

To remember again, we have 1 kg of mass attached to 5 m long spring.

At equilibrium state, the spring measure 6 m and push up and

release it from rest at a point of 50 cm above the equilibrium position.

3:16

So the corresponding characteristic

equation is r2 + 7r + 10 = 0,

which has 2 real roots, r = -5 and -2.

So that the general solution of this

homogeneous differential equation will be,

x(t) = c1e to the -5t + c2e to the -2t, right?

So this is overdamped situation, right?

Let's consider these two initial conditions and say,

x(0), from this equation, x( 0) = c1 + c2, right?

It must be equal to -0.5, okay?

And again from this equation, x'(0) that is equal

to -5c1- 2c2, and then must be equal to 0, okay?

4:22

So, solving this simultaneous equation for two unknowns,

the c1 and the c2, you will get c1 is equal to one-third and

the c2 is equal to -5 over 6, right?

And that means the solution to this initial

value problem x(t) = one-third e to

the -5t- 5 over 6 e to the -2t, right?

This is overdamped motion, okay?

To determine whether the mass passes through

the equilibrium point or not, okay?

Let's assume that the mass passes through the equilibrium position.

That means x(t) = 0, okay?

Let's determine the time at which this happens, right?

So, we set one-third e to the -5t- 5

over 6 e to the -2t = 0, okay?

So, through a simple manipulation, if you solve it then you're

going to get t =- one-third and log five-halves, right?

Because five-halves is greater than 1, log five-halves is positive.

But because of this negative sign in front,

the time t given by this quantity, this is negative, right?

What does that mean, okay?

x(t) can be 0 only at some negative time, okay?

That means in fact the mass cannot pass through the equilibrium position, okay?

6:13

As another example, let's consider the following.

In the spring mass system as in Example 2, the same situation,

find its steady state solution when there is an external

force 2 cosine 2t, acting on the system, okay?

And that means we have a non-homogeneous second order differential equation.

Say, x double prime + 7x prime +

10x = 2 cosine 2t, right?

Then by the method of undetermined coefficients,

we can see easily that there is a particular solution

xp(t) of the form A cosine 2t + B sine 2t, right?

Substitute this form of the particular solution

into the differential equation, okay?

Then you are going to get the following equations, okay?

7:21

So that means comparing the coefficients of a cosine 2t and

comparing the coefficients of sine 2t, you will get,

6A + 14B = 2 and 6B- 14A = 0, okay?

Solving this simultaneous equation,

you will get A = 3 over 58, and B = 7 over 58, right?

So that our particular solution xp(t),

that is equal to 1 over 58 times 3 cosine 2t + 7 sine 2t.

The in fact we know that this particular solution becomes

a steady state solution because the corresponding

complementary solution of this problem x sub c(t).

8:20

The complementary solution xc(t) which is a solution of

the x double prime + 7x prime + 10x = 0, okay?

We know that it tends to 0 as t tends to infinity, right?

So we know that xc(t), this is a transient solution.