*Notes to a video lecture on http://www.unizor.com*

__Viscosity Damping 2__

Critical Damping

Critical Damping

This lecture continues analyzing the damped oscillation in a case of

**critical damping**. It relies on equations and symbols presented in the previous lecture.

The main differential equation that describes the damped oscillation is

**m·x"(t) + c·x'(t) + k·x(t) = 0**where

*m*is a

**mass**of a object attached to a free end of a spring,

*c*is

**viscosity**of environment surrounding a spring, as it acts against motion of an object,

*k*is

**elasticity**of a spring,

*x(t)*is displacement of an object from its neutral position on a spring.

We also consider initial conditions

*x(0) = a*(initial stretch),

*x'(0) = 0*(no initial speed).

In search of two

*partial solutions*of the equation above we used in a previous lecture a function

**x(t) = e**^{γ·t}where

*γ*- some (generally, complex) number.

The reason for this is that, substituting this function with its first and second derivatives into the differential equation, we get

**x'(t) = γ·e**^{γ·t}

**x"(t) = γ²·e**^{γ·t}and the differential equation is

**γ²·m·e**^{γ·t}+ γ·c·e^{γ·t}+ k·e^{γ·t}= 0Canceling

*, which never equal to zero, we get a quadratic equation for*

**e**^{γ·t}*γ*

**γ²·m + γ·c + k = 0**with any of its solutions (generally, complex), used in function

*e*, delivering the solution of the original differential equation.

^{γ·t}*Case 2*

**Δ**

*=c²−4k·m**EQUALS to ZERO*

So, we assume now that the discriminant of the quadratic equation above is zero:

**Δ**.

*= c²−4k·m = 0*Then two roots to the quadratic equation above coincide:

**γ**_{1,2}= −c/(2m)As we see, in this case our approach delivered only one

*partial solution*to our differential equation:

**x**_{1}(t) = e^{−c·t/(2m)}But, to get to a

*general solution*, we need two

*partial*ones.

The second partial solution we will search among functions

*x*.

_{2}(t)=t·e^{γ·t}

**x'**_{2}(t) = e^{γ·t}+ t·γ·e^{γ·t}

**x"**_{2}(t) = γ·e^{γ·t}+ γ·e^{γ·t}+ t·γ²·e^{γ·t}and the differential equation is

**m·[γ·e**

+ c·[e^{γ·t}+ γ·e^{γ·t}+ t·γ²·e^{γ·t}] ++ c·[e

^{γ·t}+ t·γ·e^{γ·t}] + k·t·e^{γ·t}= 0Canceling exponent

*e*that never equals to zero, we obtain

^{γ·t}

**m·[γ + γ + t·γ²] +**

+ c·[1 + t·γ] + k·t = 0+ c·[1 + t·γ] + k·t = 0

Let's group the terms of this equation

**t·[m·γ² + c·γ + k] + 2m·γ + c = 0**If we can find such

*γ*that the expression above is always zero for any

*t*, the corresponding function

*x*will be a second partial solution we are looking for.

_{2}(t)=t·e^{γ·t}First of all, we should use the fact that the case we are considering now is

**Δ**.

*= c²−4k·m = 0*From this follows that

**k = c²/(4m)**Substituting this into the above equation, we get

**t·[m·γ² + c·γ + c²/(4m)] +**

+ 2m·γ + c = 0+ 2m·γ + c = 0

Now let's choose such

*γ*that nullifies the part not involved in multiplication by

*t*:

**2m·γ + c = 0**

**γ = −c/(2m)**What it does with a coefficient at

*t*is as follows:

**m·γ² + c·γ + c²/(4m) =**

= m·c²/(4m²) − c²/(2m) +

+ c²/(4m) =

= c²/(4m) − c²/(2m) +

+ c²/(4m) = 0= m·c²/(4m²) − c²/(2m) +

+ c²/(4m) =

= c²/(4m) − c²/(2m) +

+ c²/(4m) = 0

So, the coefficient at

*t*is also equals to zero.

Therefore, with

*γ=−c/(2m)*the quadratic equation above is satisfied for any

*t*and function

*x*is the second

_{2}(t)=t·e^{−c·t/(2m)}*partial*solution to our differential equations.

Now, having two

*partial*solutions

*and*

**x**_{1}(t) = e^{−c·t/(2m)}*,*

**x**_{2}(t) = t·e^{−c·t/(2m)}we can formulate a

*general solution*to

**critically damped**oscillation, when Δ

*=c²−4k·m=0*:

**x(t) = A·x**

= A·e_{1}(t) + B·x_{2}(t) == A·e

^{−c·t/(2m)}+ B·t·e^{−c·t/(2m)}We can determine coefficients

*A*and

*B*from initial conditions

*and*

**x(0) = a**

**x'(0) = 0**From this we get

*and*

**A = a**

**−A·c/(2m) + B = 0**Therefore, our two coefficients are

*A=a*and

*B=a·c/(2m)*.

General solution for of a case of

**critical damping**, when

*c²=4m·k*, is

**x(t) = a·e**

+ (a·c/(2m))·t·e

= a·(1+c·t/(2m))·e^{−c·t/(2m)}++ (a·c/(2m))·t·e

^{−c·t/(2m)}== a·(1+c·t/(2m))·e

^{−c·t/(2m)}Let's calculate the exact expression for this function for some sample values of the constants involved:

**a=12**

**m=1**

**c=10**

**k=25****Δ**

*=c²−4m·k=0*

**x(t) = 12·(1+5t)·e**^{−5t}Now let's check if this solution satisfies our differential equation

**m·x"(t) + c·x'(t) + k·x(t) = 0**Substituting the assumed values for coefficients, it looks like this:

**x"(t) + 10·x'(t) + 25·x(t) = 0**Let's check it out for our sample function

**x(t) = 12·(1+5t)·e**^{−5t}

**x'(t) = 12·5·e**

+ 12·(1+5t)·(−5)·e

= −300t·e^{−5t}++ 12·(1+5t)·(−5)·e

^{−5t}== −300t·e

^{−5t}

**x"(t) = −300·e**^{−5t}+1500t·e^{−5t}Substituting these expressions into equation

*(OK)*

**x"(t) + 10·x'(t) + 25·x(t) =**

= −300·e

− 10·300t·e

+ 25·12·(1+5t)·e

= −300·e

− 3000t·e

+ 300·e

= 0= −300·e

^{−5t}+1500t·e^{−5t}−− 10·300t·e

^{−5t}++ 25·12·(1+5t)·e

^{−5t}== −300·e

^{−5t}+ 1500t·e^{−5t}−− 3000t·e

^{−5t}++ 300·e

^{−5t}+ 1500t·e^{−5t}== 0

Let's check the initial conditions.

*(OK)*

**x(0)=12=a***(OK)*

**x'(0)=0**Let's draw a graph of our sample function

**x(t) = 12·(1+5t)·e**^{−5t}This graph confirms that the displacement of an object from the neutral position monotonically goes to zero, that is it moves in one direction only towards the neutral position

__without oscillation__.

The effect of damping of viscous environment is so strong that it prevents the object to oscillate on a spring.

This was a case of

**critical damping**.

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