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