## Wednesday, June 2, 2021

### Friction Damping 2: UNIZOR.COM - Physics4Teens - Waves - Mechanical Osci...

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

Friction Damping 2
Equation of Motion

We continue analyzing the behavior of an object of mass m oscillating on a spring of elasticity k with its oscillation damped by constant force of friction Ff=μ·m·g, where μ is the kinetic friction coefficient and g is the free fall acceleration.

Previous lecture was dealing with energy aspect of this oscillation, and we came to conclusion that on each half a cycle (movement in one particular direction) the amplitude of oscillation is diminishing by , where λ=μ·m·g/k is the critical distance parameter of an entire oscillating system.

In this lecture we will derive the law of motion - a formula describing the position of an object x(t) as a function of time t.

We assume that initially at time t=t0=0 a spring is stretched by a distance x(t0)=a significantly greater than critical distance λ to be able to follow the oscillation for a few cycles of changing the direction of motion.
We further assume, as in previous lectures, that there is no initial push of an object. After stretching the spring to position x(t0)=a we just let it go by itself. Mathematically, it means that x'(t0)=0.

For non-damped oscillation, based on the Second Newton's Law and the Hooke's Law, equating the same force acting on an object expressed according to these two laws, we have composed the main differential equation for harmonic oscillation that describes the movement of this object without a friction
m·x"(t) = −k·x(t)
Its solution, considering the initial conditions above, is
x(t) = a·cos(ω·t)
where ω = √k/m

Addition of a constant friction force Ff that acts always against the movement (that is, opposite to a velocity vector) complicates the issue, as it changes the direction after each half a cycle.

Let's consider the first half an oscillation cycle, when an object moves from the initial position x(t0)=a to a negative direction of coordinate x until it stops at some time t1 at position x(t1)=b, which we have determined in the previous lecture to be b=−(a−2λ).

The total force acting on an object is spring's elasticity Fs=−k·x(t) and friction, acting during this first half a cycle in the positive direction and equal to Ff=μ·m·g.
Therefore, the total force acting on an object as it moves from x(t0)=a to x(t1)=b is
F = −k·x(t) + μ·m·g

From the Second Newton's Law the same force equals to
F = m·x"(t)
Therefore, we have a differential equation
m·x"(t) = −k·x(t) + μ·m·g
or
x"(t) + (k/m)·x(t) − μ·g = 0
Initial conditions to this differential equation are
x(0) = a
x'(0) = 0

For convenience, we will use variable ω defined as ω²=k/m. It was useful in our discussion of harmonic oscillation and will be helpful here as well.
Then our differential equation looks like
x"(t) + ω²·x(t) − μ·g = 0

Let's try to find a solution using a cos() trigonometric function to satisfy our differential equation:
x(t) = A·cos(ω·t) + B
Then
x'(t) = −A·ω·sin(ω·t)
x"(t) = −A·ω²·cos(ω·t)

Substitute this into our differential equation:
−A·ω²·cos(ω·t) +
+ ω²·
[A·cos(ω·t) + B] − μ·g = 0
or
ω²·B = μ·g

From this we find the coefficient B (notice the using of λ, introduced above):
B = μ·g/ω² = μ·g·m/k = λ

The value of coefficient A we can find from the initial condition x(0)=a, where a is an initial stretch of a spring.
x(0) = A·cos(0) + B = A + B
Since x(0)=a and B=λ
a = A + λ
and
A = a − λ

So, the position of our object during its movement from x(t0=0)=a to x(t1)=b=−(a−2λ) is
x(t) = (a−λ)·cos(ω·t) + λ

The points where the speed of our object equals to zero are the beginning x(t0=0)=a and the end x(t1)=b of its moving in negative direction. As a checking, let's use our equation of motion to find these points.
x'(t) = −(a−λ)·ω·sin(ω·t) = 0
Since we assumed that a is significantly greater than λ, the only case when it's equal to zero is when sin(ω·t)=0, that is ω·t=π·N, where N is any integer number.

The first time (N=0) the speed is zero at the beginning, when t=t0=0 and an object is in its initial position at distance a from the neutral position.
The next one (N=1) is when ωt=π or t=t1=π/ω.
Substituting this value for time and using the property cos(π)=−1, we obtain the position where the movement stops
x(t1) = x(π/ω) =
= (a−λ)·cos(π) + λ =
= 2λ−a =
= −(a−2λ)

This is exactly the same value we obtained by using the energy considerations in the previous lecture - original deviation on stretching a, reduced by during the first half a cycle, when an object moves towards negative direction, crosses the neutral point (so, its coordinate becomes negative), squeezes a spring until it finally stops.

Let's analyze the second half of the first cycle, when an object moves from most negative position towards positive direction with expanding spring.
The initial position at this time t1=π/ω is x(t1)=−(a−2λ) with initial speed x'(t1)=0.
The differential equation describing the movement is slightly different because the force of friction now is directed towards the negative end of the axis against the direction of the object towards the positive direction.

Combined force of a spring and friction is
Fs + Ff = −k·x(t) − μ·m·g

The differential equation of motion, based on the Newton's Second Law, is
m·x"(t) = −k·x(t) − μ·m·g
or
x"(t) = −(k/m)·x(t) − μ·g

Let's try to find a solution using a cos() trigonometric function to satisfy our differential equation:
x(t) = A·cos(ω·t) + B
Then
x'(t) = −A·ω·sin(ω·t)
x"(t) = −A·ω²·cos(ω·t)

Substitute this into our differential equation:
−A·ω²·cos(ω·t) +
+ ω²·
[A·cos(ω·t) + B] + μ·g = 0
or
ω²·B = −μ·g

From this we find the coefficient B (notice the using of λ, introduced above):
B = −μ·g/ω² = −μ·g·m/k = −λ

The value of coefficient A we can find from the initial condition x(t1)=−(a−2λ), where a is an initial stretch of a spring and t1=π/ω.
x(t1) = A·cos(π) + B = −A + B
Since we have an initial condition x(t1)=−(a−2λ) and B=−λ, the equation for A is
−(a−2λ) = −A − λ
and
A = a − 3λ

Therefore, the equation of motion of an object in the second half of the first cycle is
x(t) = (a−3λ)·cos(ω·t) − λ

Very important characteristic of this motion is that its general behavior is similar to the motion during the first half of the first cycle - based on a cosine function with the same argument ω·t. It means that the time between negative most and positive most positions of an object remains constant and equal to π/ω, while amplitude is diminishing with each half a cycle by .

The speed of an object x'(t) at the end of the second half of the first cycle is zero. Let's determine the time t2 when it happens.
x'(t) = −(a−3λ)·ω·sin(ω·t) = 0
Since we assumed that a is significantly greater than λ to assure multiple cycles of an object, this expression equals to 0 when sin(ω·t) = 0, that is ω·t=π·N, where N is any integer.
N=0 corresponded to the initial position of an object at x(t0)=a,
N=1 corresponded to the position of an object at the end of the first half of the first cycle at x(t1)=−(a−2λ),
N=2 corresponds to the position of an object at the end of the second half of the first cycle.

With N=2 the time of the end of the second half of the first cycle is
t2 = 2π/ω
From this we can find the position of an object at the end of the second half of the first cycle:
x(t2 ) = (a−3λ)·cos(2π) − λ =
= a − 4λ

This corresponds to the results based on energy considerations, obtained in the previous lecture.

Summary

The motion of an object on a spring with constant friction is sinusoidal with constant timing between extreme positive and negative positions relative to the neutral. This timing equals to π/ω, where ω²=k/m - a ratio of a spring's elasticity and object's mass.

The amplitude of oscillations is diminishing on every half cycle by the same constant , where λ is a critical distance equaled to μ·m·g/k, where μ is the coefficient of kinetic friction.