Sunday, May 9, 2021

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

Notes to a video lecture on

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.
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

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
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
I·φ"(t) = −k·φ(t)
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.

Tuesday, May 4, 2021

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

Notes to a video lecture on

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·x(t)/d

(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·x(t)/dx² = −k·x(t)
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).

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

x"(t) = −C1·ω²·cos(ωt) −
− C2·ω²·sin(ωt)

x"(t) = −(k/m)·x(t)
we conclude that
−(k/m)·x(t) = −C1·ω²·cos(ωt) −
− C2·ω²·sin(ωt)

−(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.
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.