Viscosity Damping 2: UNIZOR.COM - Physics4Teens - Waves - Mechanical Osc...

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

Viscosity Damping 2
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 = 0

Canceling e^γ·t, which never equal to zero, we get a quadratic equation for γ
γ²·m + γ·c + k = 0
with any of its solutions (generally, complex), used in function e^γ·t, delivering the solution of the original differential equation.

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:
x₁(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 x₂(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^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 = 0

Canceling exponent e^γ·t that never equals to zero, we obtain
m·[γ + γ + t·γ²] +
+ 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 x₂(t)=t·e^γ·t will be a second partial solution we are looking for.

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

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
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 x₂(t)=t·e^−c·t/(2m) is the second partial solution to our differential equations.

Now, having two partial solutions
x₁(t) = e^−c·t/(2m) and
x₂(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·x₁(t) + B·x₂(t) =
= A·e^−c·t/(2m) + B·t·e^−c·t/(2m)

We can determine coefficients A and B from initial conditions
x(0) = a and
x'(0) = 0
From this we get
A = a and
−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^−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^−5t +
+ 12·(1+5t)·(−5)·e^−5t =
= −300t·e^−5t
x"(t) = −300·e^−5t+1500t·e^−5t

Substituting these expressions into equation
x"(t) + 10·x'(t) + 25·x(t) =
= −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 (OK)

Let's check the initial conditions.
x(0)=12=a (OK)
x'(0)=0 (OK)

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.

Viscosity Damping 1: UNIZOR.COM - Physics4Teens - Waves - Mechanical Osc...

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

Viscosity Damping 1
Over-Damping

Another variety of damping is related to viscosity of the environment, where oscillation takes place.
An example of this is an oscillation under water.
Each motion of an object on a spring in this case is supposed to overcome a resistance of water that surrounds it.

Interesting distinction of an oscillation under water from previously studied oscillation with friction is that the resistance of water is greater if the speed of an oscillating object is greater, while the friction resistance is constant and depends only on the weight of an object.

This property of the force of resistance to be dependent on (for our theoretical studies to be proportional to) speed is expressed by a formula that is supported by experiments:
F_v(t) = −c·x'(t)
where
t is time parameter;
F_v(t) is the force of resistance caused by viscosity of the environment where oscillation occurs;
x(t) is a displacement of an object from the neutral position on a spring (positive displacement corresponds to stretching, negative - to squeezing a spring);
x'(t) is an object's speed (first derivative of displacement);
c is a damping coefficient that depends on geometrical properties of an object and physical properties of environment, where oscillation takes place.
A minus sign in the formula signifies that the resistance force is opposite to a direction of movement, defined by a sign of a speed.

Equipped by the formula above and by Hooke's Law, describing the force of a spring on an object F_s(t) as being proportional to a displacement of an object from its neutral position on a spring x(t) with coefficient of proportionality k that depends on elasticity of a spring
F_s(t) = −k·x(t)
we come up with a total force F_t(t) acting on an object, oscillating in a viscous environment:
F_t(t) = F_s(t) + F_v(t)

Since the object's acceleration (second derivative of displacement) x"(t) and mass m are related to the total force acting on an object by the Second Newton's Law
F_t(t) = m·x"(t)
we have an equation of motion for an oscillating object in the viscous environment
m·x"(t) = −k·x(t) − c·x'(t)
or
m·x"(t) + c·x'(t) + k·x(t) = 0

This is linear differential equation that we will solve to get the formula for a displacement of an object from its neutral position on a spring x(t) as a function of time.

At this point we'd like to refer you to "Math 4 Teens" course on the same site UNIZOR.COM as this lecture will help to refresh your knowledge about linear differential equations of the second order. From the Unizor home screen choose "Math 4 Teens" - "Calculus" - "Ordinary Differential Equations, Higher Order Equations", where Hooke's Law is presented and general approach to solving linear differential equations of the second order is discussed.

The general solution to a linear differential equation of the second order, like the one above, can be found by finding two functions that satisfy this equation x₁(t) and x₂(t), called partial solutions, and describing the general solution as a linear combination of these two functions
x(t) = A·x₁(t) + B·x₂(t).
The coefficients A and B depend on two initial conditions x(0) and x'(0).

In the lecture about Hooke's Law, as an example of linear differential equation of the second order, it was suggested to seek partial solutions based on 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 = 0

Canceling e^γ·t, which never equal to zero, we get a quadratic equation for γ
γ²·m + γ·c + k = 0
with any of its solutions (generally, complex), used in function e^γ·t, delivering the solution of the original differential equation.

Since a quadratic equation has exactly two solutions among complex numbers γ₁ and γ₂, we have two solutions to our original differential equations: x₁(t)=e^γ₁·t and x₂(t)=e^γ₂·t.

Moreover, since the differential equation is linear, any linear combination of the above two solutions will be a solution to a differential equation.
Therefore, the general solution looks like
x(t) = A·e^γ₁·t + B·e^γ₂·t
where constants A and B are defined by two initial conditions of the motion described by our differential equation.

Switching from theory to concrete results for our differential equation presented above, our first task is to solve the quadratic equation to find γ₁ and γ₂
γ²·m + γ·c + k = 0

Very important for a solution to this equation is an expression under radical - discriminant Δ=c²−4k·m.
If it's not negative, both roots of our equation are real, otherwise we have to consider complex roots.

Solutions to this equations are:
γ_1,2 = [−c±√c²−4k·m] /(2m)
or
γ_1,2 = −c/(2m)±√(c/(2m))²−k/m

At this point we will consider three different cases of a discriminant of our quadratic equation:
Δ=c²−4k·m is positive
Δ=c²−4k·m equals to zero
Δ=c²−4k·m is negative
The first case is analyzed in this lecture, two others - in the next.

Case 1
Δ=c²−4k·m is POSITIVE

Let
ω² = Δ /m² = (c/(2m))²−k/m
where ω - some non-negative real number.
Obviously, ω is smaller then c/(2m).
Then the solutions to our equations are
γ_1,2 = −c/(2m) ± ω
Since ω is smaller then c/(2m), both solutions to our quadratic equation are negative.

General solution to our differential equation is
x(t) = A·e^{(−c/(2m)+ω)·t} +
+ B·e^{(−c/(2m)−ω)·t}

Let's determine coefficients A and B from initial conditions on the position and speed of an object.
At t=0 the position is x(0)=a and speed is x'(0)=0.
From this follows
A + B = a
(−c/(2m)+ω)·A +
+ (−c/(2m)−ω)·B = 0 Simplifying the second equation, we get the system of 2 linear equations with two unknowns A and B
A + B = a
−c/(2m)·(A+B) + ω·(A−B) = 0 Using the first equation for A+B=a in the second one, we obtain:
A + B = a
−c·a/(2m) + ω·(A−B) = 0
or
A + B = a
A − B = c·a/(2m·ω)

From these two equations we obtain the values of A and B:
A = a·[1+c/(2m·ω)]/2
B = a·[1−c/(2m·ω)]/2

Solution to our differential equation with given initial conditions is
x(t) = (a/2)·[1+c/(2m·ω)]·
·e^{(−c/(2m)+ω)·t} +
+ (a/2)·[1−c/(2m·ω)]·
·e^{(−c/(2m)−ω)·t}
This function describes the position of an object on a spring at any moment of time t relatively to its neutral position on a spring.

Regardless of the signs of coefficients A and B, both exponents in the above expression for x(t) are negative and each term is monotonously going to zero as time t increases to infinity, so the whole function x(t) that characterizes the displacement of the object on a spring in a viscous environment is decreasing to zero as time t increases to infinity. That is, the object moves towards its neutral position on a spring.

To further investigate the character of object's movement, let's find its speed, as it moves toward the neutral position. It requires to take the derivative of the above function describing its position.
x'(t) = (a/2)·(−c/(2m)+ω)·
·[1+c/(2m·ω)]·e^{(−c/(2m)+ω)·t} +
+ (a/2)·(−c/(2m)−ω)·
·[1−c/(2m·ω)]·e^{(−c/(2m)−ω)·t} =
= (a/2)·e^−c·t/(2m)·
·[(ω−c²/(4ωm²))·e^ωt +
+ (−ω+c²/(4ωm²))·e^−ωt] =
= (a/(8ωm²))·e^−c·t/(2m)·
·[(4ω²m²−c²)·e^ωt +
+ (−4ω²m²+c²)·e^−ωt] =
= (a/(8ωm²))·e^−c·t/(2m)·
·(−4m·k·e^ωt+4m·k·e^−ωt) =
= (a·k/(2ωm))·e^−c·t/(2m)·
·(−e^ωt+e^−ωt)

The expression for x'(t) can only be equal to zero, if e^ωt=e^−ωt,
which possible only if t=0.

Notice that, since ω is a non-negative real number smaller then c/(2m), the absolute value of x'(t) is decreasing to zero as time goes to infinity.
This means that the object's speed never equals to zero after initial moment, our object never stops, never changes the direction and monotonously moves towards its neutral position on a spring, moving slower and slower, as it comes closer to the neutral position.

Let's calculate the exact expression for this function for some sample values of the constants involved:
a=12
m=1
c=10
k=16
Δ=c²−4m·k=36
ω²=(c/(2m))²−k/m=9
ω=3
−c/(2m)+ω=−2
−c/(2m)−ω=−8
A=a·[1+c/(2m·ω)]/2=16
B=a·[1−c/(2m·ω)]/2=−4
x(t) = 16·e^−2t−4·e^−8t

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) + 16·x(t) = 0

Let's check it out for our sample function
x(t) = 16·e^−2t−4·e^−8t
x'(t) = −32·e^−2t+32·e^−8t
x"(t) = 64·e^−2t−256·e^−8t

Substituting these expressions into equation
x"(t) + 10·x'(t) + 16·x(t) =
= 64·e^−2t−256·e^−8t −
−320·e^−2t+320·e^−8t +
+ 256·e^−2t−64·e^−8t =
= (64−320+256)·e^−2t+
+ (−256+320−64)·e^−8t =
= 0 (OK)

Let's check the initial conditions.
x(0)=16−4=12=a (OK)
x'(0)=−32+32=0 (OK)

Let's draw a graph of our sample function
x(t) = 16·e^−2t−4·e^−8t


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 is called over-damping or large damping.

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 F_f=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 2λ, 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 F_f 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 F_s=−k·x(t) and friction, acting during this first half a cycle in the positive direction and equal to F_f=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 2λ 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
F_s + F_f = −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 2λ.

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 2λ, where λ is a critical distance equaled to μ·m·g/k, where μ is the coefficient of kinetic friction.

Friction Damping 1: UNIZOR.COM - Physics4Teens - Waves - Mechanical Osci...

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

Friction Damping 1
Energy Approach

In reality, unless oscillations are supported by some outside force, they gradually diminish in amplitude and, eventually, stop.
There are many reasons why it happens - some forces inside oscillating mechanism, like not ideal elasticity, external forces that prevent oscillation, like friction, viscosity of the medium surrounding the oscillating mechanism etc.

The oscillations that gradually diminish their amplitude because of one of above or similar factors are called damped oscillations.
The force that prevents the oscillations to continue indefinitely is called a damping force.
In this lecture we analyze the oscillation that experiences an external damping force of friction.

Our base oscillating mechanism is presented below. We will use it to demonstrate the damping effect of a friction against the horizontal surface, assuming our weightless spring conforms to the Hooke's Law and an object attached to it can be considered a point-mass.


Let's assume that object of mass m oscillates horizontally along the X-axis on a spring with elasticity k, while lying on a horizontal surface in the gravitational field of the Earth with free fall acceleration g and, therefore, experiences the constant force of friction as it moves left and right on the spring.
In this case there is a constant friction force acting against the movement equaled to
F_f = μ·m·g
where μ is the coefficient of friction.
For simplicity, we will not differentiate friction at motion (kinetic friction) from friction at rest (static friction) and assume they are both equal to the expression above.

Let x(t) be a horizontal displacement of the object from its neutral position on a spring, where a spring is not stretched or squeezed. When this displacement is positive, a spring is stretched, when negative, a spring is squeezed.
As in previous lectures, our initial conditions on the position and speed of an object at time t=0=t₀ are:
x(0) = a > 0 (initial stretched position of a spring with an object at distance a from the neutral position);
x'(0) = 0 (no initial speed, we just let the spring go by itself).

Since our ideal spring conforms to the Hooke's Law, which is symmetrical relative to stretching and squeezing a spring, we will examine in details only when a spring is initially stretched.

For oscillation to start, we have to assume that the force of a spring F_s stretched by an initial distance a (according to Hooke's Law, F_s=−k·a) is greater by absolute value than the friction force F_f=μ·m·g.
So, the necessary condition for a movement to start is
k·a > μ·m·g
or
a > μ·m·g/k = λ
where λ is a constant that combines the characteristics of a spring's elasticity k, object's mass m and friction μ·g, a compounded characteristic of an oscillating system as a whole.
Let's call this characteristic a system's critical distance.

Two points, x=λ and x=−λ, that are positioned exactly at critical distance from the neutral point x=0, we will call critical points with the former being a critical point of stretching and the latter - critical point of squeezing.

If, holding an object, we stretch a spring by distance a and then let it go, to start a movement, an object should be at a displacement from the neutral position greater by absolute value than critical distance λ. Otherwise, it will remain at rest, the pulling force of a spring would not be sufficient to overcome the friction force.

Let's assume that our initial spring stretch a is greater than critical distance λ, that is beyond the critical point of stretching. We expect that the pulling force of a spring will overcome friction, and an object will start moving in the negative direction of the X-axis towards the neutral position and, possibly, beyond it. As it moves, its potential energy is partially spent on friction.

Sooner or later it will stop, not reaching the point of x=−a on the opposite side from the neutral point, because, if it did, its potential energy would be the same as in the beginning of motion, which is impossible since energy is partially spent to do some work against the force of friction.

To find the stop point b, we will use the energy considerations.

When an object started its motion at time t₀=0 at position x(t₀)=a, its total (potential) energy, as we discussed in the previous lecture, was
U(t₀) = k·a²/2
When, after moving in negative direction along X-axis, it stops at coordinate b (positive or negative) at some time t₁, its total (potential) energy is
U(t₁) = k·b²/2

Since constant friction force F_f=μ·m·g was acting all the time against the object's motion, from the Law of Energy Conservation follows that the work performed by this friction force during the time of motion W_[a,b]=F_f·(a−b) equals to a difference in potential energies at the beginning and at the end of motion:
U(t₀) − U(t₁) = W_[a,b]

From this equation follows:
k·a²/2 − k·b²/2 = μ·m·g·(a−b)
(a²−b²)/2 = μ·m·g·(a−b)/k
(a−b)·(a+b)/2 = λ·(a−b)
(a+b)/2 = λ
which means that the position at the critical distance from the neutral point, that is, critical point, is a midpoint between the beginning and the end of a motion.

This is a very important observation about the role of critical points. We came up with it for initial stretching of a spring. Because the Hooke's Law is symmetrical for stretching and squeezing, exactly the same formula is true when we initially squeeze a spring with the only difference that we will use the critical point on the squeezing side x=−λ as a point of symmetry.

In other words, if we stretch a spring beyond the critical point of stretching x=λ and let it go, the object will start moving to a negative direction along the X-axis and will stop at a point symmetrical to the starting point relatively to the critical point of stretching.

Analogously, if we squeeze a spring beyond the critical point of squeezing x=−λ and let it go, the object will start moving to a positive direction along the X-axis and will stop at a point symmetrical to the starting point relatively to the critical point of squeezing.

Back to initial stretching (a>0), we can resolve the last equation for b to find exact position where our object stops its motion
a + b = 2λ
b = 2λ − a = −(a − 2λ)

Let's analyze this formula for different values of initial stretch distance a>0.

1. If initial stretch a is less than critical distance λ, new position b=2λ−a will be greater than critical distance λ. That is, we stretch a spring, let it go, and it will stretch even further by itself, which is impossible. This confirms that the motion in this case is impossible, the object will stay at rest.

2. If λ<a≤2λ, the object stops at distance 0≤b<λ from the neutral point. It's not negative, which means that an object will not go beyond the neutral point. Also, it's within the critical distance from the neutral point, which means that after stop it will not move any further, the force of a spring will not overcome friction.

3. If 2λ<a≤3λ, the object stops at distance −λ≤b<0 from the neutral point. It's negative, which means that an object will go beyond the neutral point and will squeeze a spring. Also, it's within the critical distance from the neutral point on the negative (squeezing) side, which means that after stop it will not move any further, the force of a spring will not overcome friction.

4. If 3λ<a≤4λ, the object stops at distance −2≤b<−λ from the neutral point. It's negative, which means that an object will go beyond the neutral point and will squeeze a spring. Also, it's distance from the neutral point (on the squeezing side) is greater than the critical distance, the spring is squeezed beyond the critical point of squeezing, which means that the force of a spring is greater than friction and, after stop, an object will move in the opposite (positive) direction towards the neutral position.
At this point it makes sense to consider this situation as the new beginning, and use the initial position b=2λ−a instead of a, repeating all the calculations, coming with a point x=c, where our object stops the next time, that is symmetrical to x=b relatively to the critical point of squeezing x=−λ:
(b+c)/2 = −λ
c = −2λ − b
c = −2λ − (2λ−a) = a − 4λ

The last case demonstrates that the absolute value of deviation of an object from the neutral position, the amplitude of its oscillation, is diminishing by 2λ in each direction every time.

Energy of Oscillation: UNIZOR.COM - Physics4Teens - Waves - Mechanical O...

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Energy of Oscillation

Let's consider an object of mass m on an ideal spring of elasticity k in ideal conditions (no gravity, no friction, no air resistance etc.)

What happens from the energy viewpoint, when we stretch this spring by a distance a from its neutral position?
Obviously, we supply it with some potential energy.

The object on a spring's free end will have this potential energy and, when we let a spring go, a spring will pull the object towards a neutral position, increasing its speed and, therefore, its kinetic energy.

The potential energy, meanwhile, is diminishing since the spring retracts towards its neutral position.
At the moment of crossing the neutral position an object has no potential energy, all its energy is converted into kinetic energy.

When an object moves further, squeezing a spring, it slows down, while squeezing a spring further and further, loses it kinetic energy, but increases potential energy, since a spring is squeezed more and more.

At the extreme position of a squeezed spring all the energy is again potential. An object momentarily stops at this point, having no speed and, therefore, no kinetic energy.

Then the oscillation continues in the opposite direction with similar transformation of energy from potential to kinetic and then back to potential.

Of course, total amount of energy, potential plus kinetic, should remain constant because of the Law of Energy Conservation.

Let's calculate the potential energy we give to an object on a spring by initially stretching a spring by a distance a from its neutral position.

According to the Hooke's Law, stretching a spring by an infinitesimal distance from position x to position x+dx requires a force F(x) proportional to x with a coefficient of proportionality k that depends on the properties of a spring called elasticity.

On the distance dx this force does some infinitesimal amount of work that is equal to
dW(x) = F(x)·dx = k·x·dx.
Integrating this infinitesimal amount of work on a segment from x=0 to x=a, we will obtain the total amount of work W(a) we have to spend to stretch a spring by a distance a from its neutral position.
This amount of work is the amount of potential energy U(a) we supply to an object on a stretched spring.
U(a) = ∫_[0,a]k·x·dx = k·a²/2

This formula for potential energy is true for any displacement a. This displacement can be positive (stretching) or negative (squeezing), the potential energy is always positive or zero (for a=0 at the neutral point).

Now we can easily find a speed of an object v₀ when it crosses the neutral position. This is the object's maximum speed, since potential energy at this point is zero and all energy is kinetic, which is proportional to a square of its velocity. Its kinetic energy at this point must be equal to the above value U(a). At the same time, if its speed is v₀, its kinetic energy is E₀=m·v₀²/2.
Therefore,
U(a) = m·v₀²/2
k·a²/2 = m·v₀²/2
|v₀| = √k/m·a = ω·a
where ω = √k/m is the same parameter used in expressing the harmonic oscillations in a form x(t)=a·cos(ωt).

In the above expression we used absolute value |v₀| because this speed is positive when an object moves from a squeezed position to a stretched one and negative in an opposite direction.

Using this approach we can find an object's velocity v_d at any distance d from the neutral position.
The potential energy of an object in this case is U(d)=k·d²/2.
Its kinetic energy is E(d)=m·v_d²/2.
Since the total energy is supposed to be equal the potential energy at initial position U(a)=k·a²/2, the kinetic energy equals to
E(d) = U(a) − U(d) =
= k·a²/2 − k·d²/2
From this we can find v_d:
m·v_d²/2 = k·a²/2 − k·d²/2
|v_d| = √(k/m)·(a²−d²)
|v_d| = ω·√a²−d²

The above formula for |v_d| corresponds to speed being equal to zero at the extreme position of an object at distance a from the neutral point and it being maximum at a neutral point, where d=0.

Rotational Oscillation: UNIZOR.COM - Physics4Teens - Waves - Mechanical ...

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

Rotational Oscillation

Rotational oscillations (also called torsional oscillations) can be observed in movements of a balance wheel inside hand watches. It rotates, that's why it's rotational, and it moves along the same trajectory back and forth, that's why it's oscillation.

Another example might be a weightless horizontal rod with two identical weights at its opposite ends hanging on a vertical steel wire attached to a rod's midpoint.

If we wind up the horizontal rod, as shown on the picture, and let it go, it will create a tension in the twisted wire that will start untwisting, returning the rod into its original position, then winding in an opposite direction etc., thus oscillating rotationally.

Recall the concept of a torque for rotational movement
τ = R·F
In a simple case of a force acting perpendicularly to a radius (the only case we will consider) the above can be interpreted just as a multiplication. In a more general case, assuming both force and radius are vectors, the above represents a vector product of these vectors, making a torque also a vector.

While the tension force of a twisted steel wire F₁ might be significant, it acts on a very small radius of a wire r, so the force F₂, acting on each of two weights on opposite sides of a rod of radius R and having the same torque τ, is proportionally weaker
τ = r·F₁ = R·F₂
from which follows
F₁ /F₂ = R /r
and
F₂ = (r/R)·F₁ = τ /R

Dynamics of reciprocating (back and forth) movement are expressed in terms of inertial mass m, force F and acceleration a by the Second Newton's Law
F = m·a

In case of a rotational movement with a radius of rotation R the dynamics are expressed in terms of moment of inertia I=m·R², torque τ=R·F and angular acceleration α=a/R by the rotational equivalent of the Second Newton's Law
τ = I·α

There is a rotational equivalent of a Hooke's Law. It relates a torque τ and an angular displacement φ from a neutral (untwisted) position
τ = −k·φ

For rotational oscillations an angular displacement φ is a function of time φ(t). Angular acceleration α is a second derivative of an angular displacement φ(t).

Therefore, we can equate the torque expressed according to the rotational equivalent of the Second Newton's Law to the one expressed according to the rotational equivalent of the Hooke's law, getting an equation
I·α = −k·φ
or
I·φ"(t) = −k·φ(t)
or
φ"(t) = −(k/I)·φ(t)

The differential equation above is of the same type as for an oscillations of a weight on a spring discussed in the previous lecture. The only difference is that, instead of a mass of an object m we use moment of inertia I=m·R².

For initial angular displacement (initial twist) of a steel wire φ(0)=γ and no initial angular speed (φ'(0)=0) the solution to this equation is
φ(t) = γ·cos(√k/I·t)

The rotational oscillations in our case have a period (the shortest time the object returns to its original position)
T = 2π·√I/k

Frequency of rotational oscillations f = 1/T.
Therefore, our rod with two weights makes
f = 1/T = (1/2π)·√k/I
oscillations per second.

Since I=m·R², the period is greater (and the frequency is smaller) when objects are more massive and on a greater distance from a center of a rod, where the wire is attached.

Notice that a period and a frequency of these oscillations are not dependent on initial angle of turning the rod φ(0)=γ. This parameter γ defines only the amplitude of oscillations, but not their period and frequency.

This is an important factor used, for example, in watch making with a balance wheel oscillating based on its physical characteristics and an elasticity of a spiral spring.
No matter how hard you wind a spring (or how weak it becomes after it worked for awhile), a balance wheel will maintain the same period and frequency of its oscillations.

Periodic Movement: UNIZOR.COM - Physics4Teens - Waves - Mechanical Oscil...

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Periodic Movement

Mechanics, as a subject, deals with movements of different objects. Among these movements there are those that we can call "repetitive". Examples of these repetitive movements, occurring during certain time segment, are rotation of a carousel, swinging of a pendulum, vibration of a musical tuning fork, etc. Here we are talking about certain time segment during which these movements are repetitive, because after some time these movements are changing, if left to themselves.

These repetitive movements might be of a kind when the repetitions are to a high degree exactly similar to each other (like in case of a pendulum) or some of the characteristics of the motion change in time (like in case of a tuning fork).

Repetitive movements that can be divided into equal time segments, during which the movements to a high precision repeat exactly each other, are called periodic.
The time segments of such a repetitive movement are called periods.

If a position P of an object making periodic movement with a period T is defined by a set of Cartesian coordinates P=(x,y,z) as a vector function of time P(t), the periodicity means that for any time moment t
P(t) = P(t+T)
which is exactly the mathematical definition of a periodic function.

For example, a period of rotational movement of a carousel equals to a time it takes to make one circle. The period of a movement of a pendulum is the time it moves from left most position all the way to the right most and back to the left.

A case of a vibrating tuning fork is a bit more complex because gradually the vibrations, after being initiated, diminish with time. The period of vibration might be the same during this process, but the amplitude (deviation from a middle point) would diminish with time.

A special type of periodic movement is oscillation. It's characterized by a periodic movement of an object that repeats the same trajectory of movement in alternating directions, back and forth. For example, a pendulum, an object on a spring, a tuning fork, a buoy on a surface of water under ideal weather conditions etc.

In all those systems we can observe a specific middle point position from which an object can deviate in both directions. If put initially at this position, an object would remain there, unless some external force acts on it. This is a point of a stable equilibrium. Then, after some external force is applied, it will move along its trajectory back and forth, each time passing this equilibrium point.

From this point an object can move along a trajectory to some extreme position, then back through an equilibrium point to another extreme position, then back again, repeating a movement along the same trajectory in alternating directions.

Oscillation is only possible if some external forces act on a moving object towards stable equilibrium point. Otherwise, it would never return to an equilibrium. These forces must depend on the position, not acting at the equilibrium point, acting in one direction in case an object deviated from an equilibrium to one side along its trajectory and acting in the opposite direction in case an object deviated to the other side along a trajectory.

A very important type of oscillations are so-called harmonic oscillations.
An example of this type of a movement is an object on an initially stretched (or squeezed) spring with the only force acting on an object during its movement to be the spring's elasticity.

According to the Hooke's Law, the force of elasticity of a spring is proportional to its stretch or squeeze length and directed towards a neutral point of no stretch nor squeeze.

If a string is positioned along the X-axis on a Cartesian system of coordinates with one end fixed to some point with negative coordinate on this axis, while its neutral point at x=0, the position of an object attached to this spring and oscillating can be described as a function of time x(t) that satisfies two laws:

(1) the Second Newton's Law connecting the force of elasticity F(t) to the mass m and acceleration (second derivative of position)
F(t) = m·x"(t) = m·d²x(t)/dx²

(2) the Hooke's Law connecting the force of elasticity F with a displacement of a free end of a spring from its neutral position
F(t) = −k·x(t)
(where k is a coefficient of elasticity that is a characteristic of a spring).

From these two equations we can exclude the force F(t) and get a simple differential equation that defines the position of an object at the free end of a spring x(t).
m·x"(t) = m·d²x(t)/dx² = −k·x(t)
or
x"(t) = −(k/m)·x(t)

Obviously, trigonometric functions sin(t) and cos(t) are good candidates for a solution to this equation since their second derivative looks like the original function with some coefficients
sin"(t) = -sin(t)
cos"(t) = -cos(t)

General solution to the above linear differential equation is
x(t) = C₁·cos(ωt) + C₂·sin(ωt)
where ω depends on coefficients of the differential equation and constants C₁ and C₂ depend on initial conditions (initial displacement of the object off the neutral position on a spring and its initial speed).

Then
x'(t) = −C₁·ω·sin(ωt) +
+ C₂·ω·cos(ωt)
x"(t) = −C₁·ω²·cos(ωt) −
− C₂·ω²·sin(ωt)

Since
x"(t) = −(k/m)·x(t)
we conclude that
−(k/m)·x(t) = −C₁·ω²·cos(ωt) −
− C₂·ω²·sin(ωt)
or
−(k/m)·[C₁·cos(ωt) +
+ C₂·sin(ωt)] =
−C₁·ω²·cos(ωt) −
− C₂·ω²·sin(ωt)
from which immediately follows
ω = √k/m

Assume, initially we stretch a spring by a distance a from the neutral position (that is, x(0)=a) and let it go without any push (that is, x'(0)=0).
From these initial conditions we can derive the values of constants C₁ and C₂
a = x(0) =
= C₁·cos(0) + C₂·sin(0) = C₁
0 = x'(0) =
= −C₁·ω·sin(0) + C₂·ω·cos(0) =
= C₂·ω
from which immediately follows
C₁ = a
C₂ = 0
and the solution for our differential equation with given initial conditions is
x(t) = a·cos(√k/m·t)

The oscillations described by the above function x(t) in its general form x(t)=a·cos(ω·t) are called simple harmonic oscillations.

Parameter a characterizes the amplitude of harmonic oscillations, while parameter ω represents the angular speed of oscillations.
Function cos(t) is periodical with a period T=2π.
Function cos(ωt) is also periodical with a period T=2π/ω.
Indeed,
cos(ω(t+T)) =
= cos(ω(t+2π/ω)) =
= cos(ωt+2π) =
= cos(ωt)
Therefore, the simple harmonic oscillations in our case have a period (the shortest time the object returns to its original position)
T = 2π/ω = 2π·√m/k

If one full cycle the oscillation process makes in time T, we can find how many cycles it makes in a unit of time (1 sec) using a simple proportion
1 cycle - T sec
f cycles - 1 sec
Hence, f = 1/T
Therefore, the object on a spring we deal with makes
f = 1/T = (1/2π)·√k/m
oscillations per second.