Monday, July 19, 2021

Waves in Medium: UNIZOR.COM - Physics4Teens - Waves - Waves in Medium

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

Waves in Medium

This lecture is about a different type of waves.
These waves are still mechanical oscillations in a sense that they are movements of objects with some periodicity, but the objects are not exactly the same as we studied in a previous chapter "Mechanical Oscillations".
The objects we discuss in this lecture are molecules of some substance, like air or water, or steel.

The word medium we use in a sense of "a substance regarded as the means of transmission of a force or effect", as defined in Merriam-Webster dictionary. Examples are air, water, wood, glass etc.
Basically, any substance can be a medium and transmit an oscillation from its source to some destination. The only thing that cannot be a medium in this sense is an absolute vacuum void of any force fields.

As an example of oscillation of a medium, consider a ringing bell after we strike it with a hammer.
For quite some time molecules inside this bell are oscillating, forcing to oscillate air around a bell as well, so we hear the sound.

Let's create a simplified model of this process.
First of all, we will research only a one-dimensional case.
As an example, consider a propagation of oscillation inside a thin metal rod after we strike one of its ends with a hammer.

Further consider that each molecule of a metal this thin rod is made of is connected by spring-like inter-molecular links with its left and right neighbors along a rod. If one molecule is oscillates, it will squeeze one link that attaches it to one neighboring molecule and stretches the other link, thus engaging its neighbors in oscillation.
In our one-dimensional case this model looks like this

where letter m indicates a molecule with its mass, while letter k indicates an elasticity of a weightless spring-like connection between molecules.

The above model of propagation of oscillation, while not perfect, allows to better understand the mechanism of waves in medium.

Finally, we simplify the problem even further by limiting the length of a rod to only two molecules, thus making the following model.



In this model two identical molecules, #1 (left on the picture) and #2 (right), of mass m each are linked by identical weightless springs of length L and elasticity k among themselves and to the edges of a thin rod.
Assume that initially the left end of the rod is at the origin of X-axis, the molecule #1 is at distance L from it along the X-axis, the molecule #2 is at distance L from the first further along the X-axis and the right end of a rod is by L further down. This assumption means that initially all springs are in neutral position.
Also assume that the molecule #1 (left on the picture above) has an initial speed v, modeling a strike of a hammer on the bell.

Let x1(t) be a displacement of the molecule #1 (left on the picture above) from its initial position at moment in time t and x2(t) be a displacement of the molecule #2 (right on the picture) from its initial neutral position at time t.
Assume, the initial displacement of molecules at time t=0 is:
x1(0)=0
x2(0)=0
Their initial velocity is:
x'1(0)=v
x'2(0)=0

There are two forces acting on the molecule #1 (left on the picture above) from two springs on both sides of it, which are:
F1L(t) by the left spring attached to the left edge and
F12(t) by the middle spring attached to the molecule #2.
There are two forces acting on the molecule #2 (right) from two springs on both sides of it, which are:
F2R(t) by the right spring attached to the right edge and
F21(t) by the middle spring attached to the molecule #1.

The direction (sign) of these forces depends on the displacement of molecules, we will take it into account when constructing the equations based on the Hooke's Law.
Also, forces F12(t) by the middle spring on the first molecule and F21(t) by the middle spring on the second molecule are opposite in direction and equal by absolute value since they depend only on the length of the middle spring.

The resulting force acting on the molecule #1 is
F1(t) = F1L(t) + F12(t)
The resulting force acting on the molecule #2 is
F2(t) = F2R(t) + F21(t)

If the displacements of molecules #1 (left on the picture above) and #2 (right) at any moment of time t are, correspondingly, x1(t) and x2(t), the forces F1L(t) of the left spring and F2R(t) of the right spring, acting on molecules #1 and #2 correspondingly, according to the Hooke's Law, are
F1L(t) = −k·x1(t)
F2R(t) = −k·x2(t)

The forces of the middle spring acting on molecules #1 and #2 depend on the difference in displacements of these molecules from their neutral position.

Let
Δ(t) = x1(t) − x2(t)

If Δ(t) is positive, the middle spring is squeezed and its left end (towards molecule #1) pushes molecule #1 to the left (in negative direction), which makes F12(t) negative, while its right end pushes to the right (in positive direction) molecule #2, which makes F21(t) positive.

If Δ(t) is negative, the middle spring is stretched and its left end (towards molecule #1) pulls molecule #1 to the right (in positive direction), which makes F12(t) positive, while its right end pulls to the left (in negative direction) molecule #2, making F21(t) negative.

Therefore,
F12(t) = −k·Δ(t)
F21(t) = k·Δ(t)

Now we can express the resulting force acting on the molecule #1 is
F1(t) = −k·x1(t) − k·Δ(t) =
= −k·
[x1(t) + x1(t) − x2(t)]
And the resulting force acting on the molecule #2 is
F2(t) = −k·x2(t) + k·Δ(t) =
= −k·
[x2(t) − x1(t) + x2(t)]

Taking into account the Second Newton's Law, F1(t)=m·x"1(t) and F2(t)=m·x"2(t), we obtain the following system of two linear homogeneous differential equations of second order
m·x"1(t) = −2k·x1(t) + k·x2(t)
m·x"2(t) = k·x1(t) − 2k·x2(t)
Bringing all members to the left part of equations, we get m·x"1(t) + 2k·x1(t) − k·x2(t) = 0
m·x"2(t) − k·x1(t) + 2k·x2(t) = 0
or, using the familiar concept of angular speed ω²=k/m,
x"1(t) + 2ω²·x1(t) − ω²·x2(t) = 0
x"2(t) − ω²·x1(t) + 2ω²·x2(t) = 0

Let's approach this system of equations in a more "mathematical" way, using vectors and matrices (you can refer to "Math 4 Teens" course on UNIZOR.COM, which explains the concepts of a vector and a matrix and operations with them).

Consider a vector X(t) of displacements
x1(t)
x2(t)
and a matrix Ω of coefficients
2ω²−ω²
−ω²2ω²

Now our system of two differential equations can be represented as one equation with vector argument
X"(t) + Ω·X(t) = 0
where X"(t) signifies a vector of second derivatives of each component of a vector X(t) and 0 signifies a null-vector.
An operation · (dot) in this equation signifies a multiplication of a matrix Ω by a vector X(t).

Compare it with an equation for a periodic movement we discussed in the very first lecture on Waves
x"(t) + (k/m)·x(t) = 0
where later on we assigned ω²=k/m.
We will try to solve a vector equation the same way we solved this similar equation in a one-dimensional case.
This is a subject of the next lecture.

Saturday, July 3, 2021

Forced Oscillation 2 - Resonance: UNIZOR.COM - Physics4Teens - Waves - M...

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

Forced Oscillation 2 -
Resonance


This lecture continues analysis of forced oscillations for a specific case, when the angular frequency of external periodic force equals to an inherent natural frequency of an object on a spring without any external forces.

Let us remind the introductory material from the previous lecture.

Oscillation is forced, when some periodic force is applied to an object that can potentially oscillate.

A subject of our analysis will be an object on a spring with no other forces acting on it except one external force that we assume is periodical.

As we know, in the absence of external force the differential equation describing the displacement x(t) from a neutral position of an object of mass m on a spring of elasticity k follows from the Hooke's Law and the Newton's Second Law
m·x"(t) = −k·x(t)
or
m·x"(t) + k·x(t) = 0
with a general solution
x(t) = A·cos(ω0·t) + B·sin(ω0·t)
where
A and B are any constants;
ω0 = √k/m is an inherent natural angular frequency of oscillation of this particular object on this particular spring.

The equation above is homogeneous because, together with any its solution x(t), the function C·x(t) will be a solution as well.

Consider a model of forced oscillation of an object on a spring with a periodic external force F(t)=F0·cos(ω·t).

Previous lecture was dedicated to a case of angular frequency of an external force ω to be different from the inherent natural frequency of an object on a spring ω0.
In this lecture we consider a case of angular frequency ω of the periodic external function F(t)=F0·cos(ω·t) to be equal to the inherent natural angular frequency ω0=√k/m of oscillations without external forces:
ω = ω0 = √k/m
So, from now on we will use only symbol ω for both inherent natural angular frequency of a spring with an object without external forces and for an angular frequency of an external force.

If external force F(t), described above, is present, the differential equation that describes the oscillation of an object looks like
m·x"(t) = −k·x(t) + F(t)
or, assuming the external force is a periodic function F(t)=F0·cos(ω·t),
m·x"(t)+ k·x(t) = F0·cos(ω·t)
where ω=√k/m

The function on the right of this equation makes this equation non-homogeneous.
Our task is to analyze the movement of an object on a spring in the presence of a periodic external function, as described by the above differential equations and some initial conditions.

Notice that if some non-homogeneous linear differential equation has two partial solutions x1(t) and x2(t), their difference is a partial solution to a corresponding homogeneous linear differential equation.
Indeed, if
m·x1"(t)+ k·x1(t) = F0·cos(ω·t)
and
m·x2"(t)+ k·x2(t) = F0·cos(ω·t)
then for x3(t)=x1(t)−x2(t)
m·x3"(t)+ k·x3(t) = 0

From the above observation follows that, in order to find a general solution to a non-homogeneous linear differential equation, it is sufficient to find its one partial solution and add to it a general solution of a corresponding homogeneous equation.

In our case we already know the general solution to a corresponding homogeneous equation
m·x"(t)+ k·x(t) = 0
So, all we need is to find a single partial solution to
m·x"(t)+ k·x(t) = F0·cos(ω·t)
where ω=√k/m
and add to it the general solution to the above homogeneous equation.

In the previous lecture, where we considered an angular frequency of external force ω to be different from the inherent natural frequency of oscillation of an object on a spring ω0, we were looking for a partial solution in a form xp(t)=C·cos(ω·t) and found such a coefficient C that the equation is satisfied:
C = F0 / [m·(ω0²−ω²)]
But, if ω=ω0=√k/m, we cannot use this method (since denominator becomes zero) and have to invent something new.

The usual recommendation in a case like this is to still use trigonometric functions, but with some simple function as a multiplier.
Let's try the following function:
xp(t) = Q·t·sin(ω·t)

Let's find second derivative of xp(t), substitute xp(t) and x"p(t) into our differential equation and find suitable values for a constant Q. That would deliver a partial solution to our non-homogeneous differential equation.

x'p(t) =
= Q·sin(ω·t) + Q·t·ω·cos(ω·t)


x"p(t) = Q·ω·cos(ω·t)+
+Q·ω·cos(ω·t)−Q·t·ω²·sin(ω·t)


Our original equation
m·x"(t)+ k·x(t) = F0·cos(ω·t)
can be written as
x"(t) + (k/m)·x(t) =
= (F0/m)·cos(ω·t)

or
x"(t)+ω²·x(t) = (F0/m)·cos(ω·t)

Substitute the partial solution expression into the left part of the differential equation above:
Q·ω·cos(ω·t)+Q·ω·cos(ω·t)−
−Q·t·ω²·sin(ω·t)+Q·t·ω²·sin(ω·t)


Canceling positive and negative terms that are equal by absolute value, got the following equation with the proposed solution
2Q·ω·cos(ω·t) =
= (F0/m)·cos(ω·t)

from which
Q = F0/(2m·ω)
The partial solution to our non-homogeneous differential equation is
xp(t) = F0·t·sin(ω·t)/(2m·ω)

Now we can express the general solution to a differential equation that describes the forced oscillation by a periodic external function as
x(t) = F0·t·sin(ω·t)/(2m·ω) +
+ A·cos(ω·t) + B·sin(ω·t)

where
A and B are any constants determined by initial conditions x(0) and x'(0);
ω = √k/m is an inherent natural angular frequency of oscillation of this particular object on this particular spring;
F0·cos(ω·t) is an external periodic force acting on an object with the same angular frequency as an inherent natural angular frequency of oscillation of this particular object on this particular spring.

It's beneficial to represent A·cos(ω·t)+B·sin(ω·t) through a single trigonometric function as follows.
Let
D = √A²+B²
Angle φ is defined by
cos(φ) = A/D
sin(φ) = B/D
(on a Cartesian coordinate plane this angle is the one from the X-axis to a vector with coordinates {A,B})
Then
A·cos(ω·t)+B·sin(ω·t) =
= D·
[cos(φ)·cos(ω·t) +
+ sin(φ)·sin(ω·t)
] =
= D·cos(ω·t−φ)


Now the general solution to our differential equation that describes forced oscillation looks like
x(t)=F0·t·sin(ω·t)/(2m·ω) +
+ D·cos(ω·t−φ)

where
F0 is an amplitude of a periodic external force;
m - mass of an object on a spring;
k - elasticity of a spring;
ω=√k/m is an inherent natural angular frequency of oscillation of this particular object on this particular spring and the angular frequency of external periodic force;
t is time;
D and φ parameters are determined by initial conditions for x(0) and x'(0).

The main component of the function describing the movement of an object, when a periodic external force acts synchronously with the inherent natural oscillation of an object on a spring, is F0·t·sin(ω·t)/(2m·ω).
This function represents oscillation with gradually, proportionally to time t, increasing to infinity amplitude.
Added to this function is a regular sinusoidal oscillation with the same angular frequency but, potentially, a phase shift φ.
The periodic external force, acting synchronously (that is, with the same angular frequency) with inherent natural oscillation of an object on a spring, causes proportional to time increasing of an amplitude of oscillations to (theoretically) infinity. This is called the resonance.

Let's represent it graphically for some specific case of initial conditions and parameters.

Initial conditions:
x(0) = 0
x'(0) = 0
From these initial conditions follows
0 = D·cos(φ)
0 = D·ω·sin(φ)
We know that ω≠0.
Also, sin(φ) and cos(φ) cannot be simultaneously equal to 0.
Therefore, D=0, and our oscillation in this case is described by a function
x(t) = F0·t·sin(ω·t)/(2m·ω)

Parameters:
m=1
k=4
ω=√k/m=2
F0=12

With the above initial conditions and parameters the oscillation of the object is described by function
x(t)=12·t·sin(2t)/(2·1·2) = 3·t·sin(2t)

The graph of this function looks like this

Tuesday, June 29, 2021

Forced Oscillation 1: UNIZOR.COM - Physics4Teens - Waves - Mechanical Os...

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

Forced Oscillation 1

Oscillation is forced, when some periodic force is applied to an object that can potentially oscillate.

A subject of our analysis will be an object on a spring with no other forces acting on it except one external force that we assume is periodical.

As we know, in the absence of external force the differential equation describing the movement of an object on a spring follows from the Hooke's Law and the Newton's Second Law
m·x"(t) = −k·x(t)
or
m·x"(t) + k·x(t) = 0
with a general solution
x(t) = A·cos(ω0·t) + B·sin(ω0·t)
where
A and B are any constants;
ω0 = √k/m is an inherent natural angular frequency of oscillation of this particular object on this particular spring.

The equation above is homogeneous because, together with any its solution x(t), the function C·x(t) will be a solution as well.

Consider a model of forced oscillation of an object on a spring with a periodic external force acting on it with the following characteristics:
elasticity of a spring k,
mass of an object is m,
force applied to an object is F(t)=F0·cos(ω·t).

In this lecture we consider a case of angular frequency ω of the periodic external function F(t)=F0·cos(ω·t) to be different from the inherent natural angular frequency ω0=√k/m of oscillations without external forces:
ω ≠ ω0 = √k/m

If external force F(t), described above, is present, the differential equation that describes the oscillation of an object looks like
m·x"(t) = −k·x(t) + F(t)
or, assuming the external force is a periodic function F(t)=F0·cos(ω·t),
m·x"(t)+ k·x(t) = F0·cos(ω·t)
The function on the right of this equation makes this equation non-homogeneous.
Our task is to analyze the movement of an object on a spring in the presence of a periodic external function, as described by the above differential equations and initial conditions.

Notice that if some non-homogeneous linear differential equation has two partial solutions x1(t) and x2(t), their difference is a partial solution to a corresponding homogeneous linear differential equation.
Indeed, if
m·x1"(t)+ k·x1(t) = F0·cos(ω·t)
and
m·x2"(t)+ k·x2(t) = F0·cos(ω·t)
then for x3(t)=x1(t)−x2(t)
m·x3"(t)+ k·x3(t) = 0

From the above observation follows that, in order to find a general solution to a non-homogeneous linear differential equation, it is sufficient to find its one partial solution and add to it a general solution of a corresponding homogeneous equation.

In our case we already know the general solution to a corresponding homogeneous equation
m·x"(t)+ k·x(t) = 0
So, all we need is to find a single partial solution to
m·x"(t)+ k·x(t) = F0·cos(ω·t)
and add to it the general solution to the above homogeneous equation.

Considering our differential equation, that describes the forced oscillation with a periodic external function, contains function x(t), its second derivative and function cos(ω·t) on its right side, it's reasonable to look for a partial solution xp(t) in a form of xp(t)=C·cos(ω·t) and find such a coefficient C that the equation is satisfied.
Then
xp(t) = C·cos(ω·t)
x'p(t) = −C·ω·sin(ω·t)
x"p(t) = −C·ω²·cos(ω·t)
Substituting these into our differential equation, we get
−m·C·ω²·cos(ω·t) +
+ k·C·cos(ω·t) = F0·cos(ω·t)


Since this is supposed to be true for any time value t, the coefficients at cos(ω·t) must satisfy the equation:
−m·C·ω² + k·C = F0
from which follows
C = F0 /(k−m·ω²) =
= F0 /m·((k/m)−ω²) =
= F0 /
[m·(ω0²−ω²)]

We have found a partial solution to our non-homogeneous equation:
xp(t)=F0·cos(ω·t)/[m·(ω0²−ω²)]

Now we can express the general solution to a differential equation that describes the forced oscillation by a periodic external function as
x(t)=F0·cos(ω·t)/[m(ω0²−ω²)] +
+ A·cos(ω0·t) + B·sin(ω0·t)

where
A and B are any constants determined by initial conditions x(0) and x'(0);
ω0 = √k/m is an inherent natural angular frequency of oscillation of this particular object on this particular spring;
F0·cos(ω·t) is an external periodic force acting on an object.

It's beneficial to represent A·cos(ω0·t)+B·sin(ω0·t) through a single trigonometric function as follows.
Let
D = √A²+B²
Angle φ is defined by
cos(φ) = A/D
sin(φ) = B/D
(on a Cartesian coordinate plane this angle is the one from the X-axis to a vector with coordinates {A,B})
Then
A·cos(ω0·t)+B·sin(ω0·t) =
= D·
[cos(φ)·cos(ω0·t) +
+ sin(φ)·sin(ω0·t)
] =
= D·cos(ω0·t−φ)


Now the general solution to our differential equation that describes forced oscillation looks like
x(t)=F0·cos(ω·t)/[m(ω0²−ω²)] +
+ D·cos(ω0·t−φ)

where
F0 is an amplitude of a periodic external force;
ω - angular frequency of a periodic external force;
m - mass of an object on a spring;
k - elasticity of a spring;
ω0=√k/m is an inherent natural angular frequency of oscillation of this particular object on this particular spring;
t is time;
D and φ parameters are determined by initial conditions for x(0) and x'(0).

As you see from the formula for a general solution, it's essential that ω0≠ω, that is an angular frequency of the external force is not equal to an inherent natural angular frequency of the object and a spring in the absence of external force.
Their equality renders zero in the denominator of a formula, invalidating the whole approach.
What happens if these frequencies are equal is a subject of the next lecture.

The solution to a forced oscillation in a case of ω0≠ω represents a superposition of two sinusoidal functions with different frequencies.
Let's represent it graphically for some specific case of initial conditions and parameters.

Initial conditions:
x(0) = 0
x'(0) = 0
From these initial conditions follows
0 = F0/[m(ω0²−ω²)] + D·cos(φ)
0 = D·ω0·sin(φ)
The second equation results in φ=0, from which the first equation gives
D = −F0/[m(ω0²−ω²)]
Parameters:
m=1
k=4
ω0=√k/m=2
F0=3
ω=5
The parameters above define the value of D:
D=−3/(4−25)=1/7

With the above initial conditions and parameters the oscillation of the object is described by function
x(t)=−3·cos(5t)/21+cos(2t)/7 =
=
[cos(2t)−cos(5t)]/7

The graph of this function looks like this

Saturday, June 26, 2021

Viscosity Damping 3: UNIZOR.COM - Physics4Teens - Waves - Mechanical Osc...

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

Viscosity Damping 3
Underdamping


This lecture continues analyzing the damped oscillation in a case of underdamping. It relies on equations and symbols presented in the previous two lectures.

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 a function
x(t) = eγ·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·eγ·t
x"(t) = γ²·eγ·t
and the differential equation is
γ²·m·eγ·t + γ·c·eγ·t + k·eγ·t = 0

Canceling eγ·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 eγ·t, delivering the partial solution of the original differential equation.

Case 3

Discriminant of this quadratic equation Δ=c²−4k·m is NEGATIVE

Let
ω² = −Δ /4m² = k/m − (c/(2m))²
where ω - some positive real number.

Then the solution to a quadratic equation for γ above is a set of two complex numbers
γ1,2 = −c/(2m) ± ω·i
where i²=−1 is an imaginary unit.

Two partial solutions of our differential equation in the domain of complex functions of real argument t are
x1(t) = e [−c/(2m)+ω·i]·t and
x2(t) = e [−c/(2m)−ω·i]·t

Recall the Euler's formula
ei·t = cos(t) + i·sin(t)
that you can find in the "Math 4 Teens" course on this site UNIZOR.COM - Math 4 Teens - Trigonometry - Complex Numbers and Trigonometry - Euler's Formula.

Using Euler's formula, we can split two partial solutions of our differential equation into real and imaginary parts:
x1(t) = x1.Re(t) + i·x1.Im(t)
where
x1.Re(t) = e−c·t/(2m)·cos(ω·t)
and
x1.Im(t) = e−c·t/(2m)·sin(ω·t)
Similarly,
x2(t) = x2.Re(t) + i·x2.Im(t)
where
x2.Re(t) = e−c·t/(2m)·cos(−ω·t) =
= e−c·t/(2m)·cos(ω·t)

and
x2.Im(t) = e−c·t/(2m)·sin(−ω·t) =
= e−c·t/(2m)·sin(ω·t)


If a complex function x(t) satisfies a linear differential equation, its real and imaginary parts must separately satisfy it as well. This obvious statement follows from the fact that, if a(t)+i·b(t)=0 for all t, then a(t)=0 and b(t)=0 for all t.

Therefore, if
x(t) = xRe(t) + i·xIm(t)
is a solution of our differential equation, then
xRe(t) and xIm(t)
are solutions in their own rights.

Therefore, we have two linearly independent real functions that are partial solutions to our differential equation:
x1(t) = e−c·t/(2m)·cos(ω·t)
x2(t) = e−c·t/(2m)·sin(ω·t)

Any linear combination of these two functions also is a solution to our differential equation.
This allows to express a general solution as
x(t) = e−c·t/(2m) ·
·
[C1·cos(ω·t)+C2·sin(ω·t)]

The expression above for any C1 and C2 not simultaneously equal to zero can be simplified as follows.
Let φ be such an angle that
cos(φ) = C1 /C1²+C2²
sin(φ) = C2 /C1²+C2²
This angle can easily be constructed, it's an angle from the X-axis to a vector with endpoint at coordinates (C1,C2).
Then
x(t) = e−c·t/(2m)·√C1²+C2²·
·
[cos(φ)·cos(ω·t) +
+ sin(φ)·sin(ω·t)
] =
= e−c·t/(2m)·√C1²+C2² ·
· cos(ω·t−φ) =
= D·e−c·t/(2m)·cos(ω·t−φ)

where D=√C1²+C2² can be any positive real number not equal to zero.
This is a general solution of our differential equation in the case of negative discriminant Δ=c²−4m·k,
where
D - any positive constant
φ - any angle (phase shift)
ω= √−Δ /(2m).

Simple check confirms that this is a solution to our differential equation.

As we see, the motion is a combination of sinusoidal oscillation cos(ω·t−φ) and a damping factor D·e−c·t/(2m), which reduces the amplitude of oscillation, as the time goes by.
The amplitude of the oscillation is exponentially diminishing
.

Let's determine constants D and φ 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.
x(t) = D·e−c·t/(2m)·cos(ω·t−φ)
x'(t) = −D·e−c·t/(2m) ·
·
[ω·sin(ω·t−φ) +
+ (c/(2m))·cos(ω·t−φ)
]

From this for t=0 follows
D·cos(−φ) = a
−D·[ω·sin(−φ) +
+ (c/(2m))·cos(−φ)
] = 0
or
D·cos(φ) = a
[−ω·sin(φ) +
+ (c/(2m))·cos(φ)
] = 0
or
D·cos(φ) = a
−D·ω·sin(φ) + c·a/(2m) = 0
or
D·cos(φ) = a
D·sin(φ) = c·a/(2m·ω)

From these we can obtain
D² = a²·[1+c²/(4m²·ω²)]
D = a·√1+c²/(4m²·ω²)

For phase φ we can find
cos(φ) =
= 1/√1+c²/(4m²·ω²)

sin(φ) =
= c/[2m·ω·√1+c²/(4m²·ω²)]


Let's calculate the exact expression for this function for some sample values of the constants involved:
a=√3
m=1
c=2
k=4
Δ=c²−4m·k=−12
ω²=−(c/(2m))²+k/m=3
D=a·√1+c²/(4m²·ω²)=2
cos(φ)=
=1/√1+c²/(4m²·ω²)=√3/2

sin(φ)=
=c/2m·ω·√1+c²/(4m²·ω²)=1/2

φ=π/6

Finally, the equation that describes an object's motion
x(t) = D·e−c·t/(2m)·cos(ω·t−φ)=
= 2·e−t·cos(√3·t−π/6)


Let's draw a graph of our sample function
x(t) = 2·e−t·cos(√3·t−π/6)


This graph confirms that the real oscillation does take place in case Δ=c²−4k·m is NEGATIVE, but its amplitude is exponentially diminishes due to damping effect.
The effect of damping of viscous environment is not strong enough to completely suppress the back and forth movement of an object on a spring, but sufficient to make these oscillation less and less significant.
This is called underdamping or small damping.

Sunday, June 20, 2021

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) = eγ·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·eγ·t
x"(t) = γ²·eγ·t
and the differential equation is
γ²·m·eγ·t + γ·c·eγ·t + k·eγ·t = 0

Canceling eγ·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 eγ·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:
γ1,2 = −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·eγ·t.
x'2(t) = eγ·t + t·γ·eγ·t
x"2(t) = γ·eγ·t + γ·eγ·t + t·γ²·eγ·t
and the differential equation is
m·[γ·eγ·t + γ·eγ·t + t·γ²·eγ·t] +
+ c·[eγ·t + t·γ·eγ·t] + k·t·eγ·t = 0


Canceling exponent eγ·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·eγ·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.

Saturday, June 19, 2021

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:
Fv(t) = −c·x'(t)
where
t is time parameter;
Fv(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 Fs(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
Fs(t) = −k·x(t)
we come up with a total force Ft(t) acting on an object, oscillating in a viscous environment:
Ft(t) = Fs(t) + Fv(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
Ft(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) = eγ·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·eγ·t
x"(t) = γ²·eγ·t
and the differential equation is
γ²·m·eγ·t + γ·c·eγ·t + k·eγ·t = 0

Canceling eγ·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 eγ·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)=eγ1·t and x2(t)=eγ2·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·eγ1·t + B·eγ2·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:
γ1,2 = [−c±√c²−4k·m] /(2m)
or
γ1,2 = −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
γ1,2 = −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))·eωt +
+ (−ω+c²/(4ωm²))·e−ωt
] =
= (a/(8ωm²))·e−c·t/(2m)·
·
[(4ω²m²−c²)·eωt +
+ (−4ω²m²+c²)·e−ωt
] =
= (a/(8ωm²))·e−c·t/(2m)·
·(−4m·k·eωt+4m·k·e−ωt) =
= (a·k/(2ωm))·e−c·t/(2m)·
·(−eωt+e−ωt)


The expression for x'(t) can only be equal to zero, if eωt=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.

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.

Sunday, May 23, 2021

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
Ff = 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 Fs stretched by an initial distance a (according to Hooke's Law, Fs=−k·a) is greater by absolute value than the friction force Ff=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 Ff=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]=Ff·(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 in each direction every time.

Sunday, May 9, 2021

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

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

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 vd 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·vd²/2.
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 vd:
m·vd²/2 = k·a²/2 − k·d²/2
|vd| = √(k/m)·(a²−d²)
|vd| = ω·√a²−d²

The above formula for |vd| 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.

Friday, May 7, 2021

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.