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Файл:A General Course of Physics. Mechanics. Textbook
.pdf
81
The revolution rate of Earth along the solar orbit is 29 km/s, the
3
v
tangential velocity of the Earth’s surface at the equator determined by the
Earth’s daily rotation is 0.47 km/s. The third cosmic velocity equals
= 13.5 km/s if the orbital velocity of the Earth’s motion and the tan-
gential velocity of the Earth’s surface motion at the equator are used as
launch velocities.
Test Questions
1. Jupiter’s orbital circulation period around the Sun is 12 times
longer than the corresponding Earth’s period. Considering the
planets’ orbits to be circular, determine what fold the distance
from Jupiter to the Sun exceeds that from the Earth to the Sun.
2. At what height above the Earth’s surface is gravitational acceleration half as much as gravitational acceleration near the
Earth’s surface? (the Earth’s radius is RE = 6370 km)
3. The mass of a rocket moving along a circular trajectory around
the Earth is 10 tonnes (m = 10 t). The rocket moves at a height
of 200 km above the Earth’s surface. Determine the potential
energy that the rocket possesses with respect to the Earth’s sur-
face.
What amount of energy is it necessary to pass to a satellite so
that it can escape the Earth’s gravity and become an independent planet revolving along the same orbit as the Earth?
4. Find the Earth’s linear velocity in its orbit assuming that the
Sun’s mass is 2-1030 kg, the distance from the Earth to the Sun
is 1.5-108 km and the Earth’s orbit is circular.
5. The planet Mars has two satellites: Phobos and Deimos. The
former is at a distance of R = 9500 km away from Mars’s center, the latter is at a distance of R = 24000 km. Find the orbital
periods of these satellites around Mars. (The mass of Mars is
0.108 that of the Earth. Consider the satellites’ orbits to be circular.)
82
6. At what distance away from the Earth’s center and at what ve-
1
v
2
v
locity should a satellite move along a circular trajectory being
in the equator plane so that it stayed at the same point above
the Earth’s surface? (The Earth’s mass is МE = 5.9 · 1024 kg)
Such satellites, called geosynchronous, are used for transmitting phone and television signals.
7. The orbital period of a binary star is 2 years. The velocities of
its constituents are
that the star moves in a circular orbit, calculate the masses of
the constituents and the distance between the constituents of
the binary star (velocities are determined according to the
Doppler effect for spectral lines of an optical spectrum; the period – according to ellipses).
8. Determine the gravity force of a body and its weight if the mass
of the body is 50 kg (m = 50) and its acceleration is а = 3 m/s2,
provided that the body is moving vertically downwards near
the Earth’s surface.
= 10 km/s and
= 5 km/s. Assuming
83
6. OSCILLATORY MOTION
kxF
maF
x
dt
xd
a
2
2
xmkx
x
m
k
x
6.1. Harmonic Oscillations
Oscillation, or oscillatory motion, is a repeated motion in which
an object repeats the same movement over and over with time.
Inherently, oscillations are quite diverse: mechanical, electro-
magnetic, electromechanical, etc.
Oscillations are called periodic if the values of physical variables
which change in the process of oscillation repeat at regular intervals.
The simplest and the most important type of periodic oscillations
are harmonic oscillations which are produced in conformity with the sine
or cosine law. In particular, harmonic oscillations are created under the
action of elasticity. Forces, which are non-elastic by their nature but similar to elastic forces in view of their dependence on displacement, are
called quasi-elastic. Oscillations, produced under the action of quasielastic forces, are also harmonic oscillations.
Let us derive the equation of harmonic oscillatory motion.
The elastic, or quasi-elastic, force F is expressed as follows:
, (58)
viz. x is the displacement of the oscillating point from its equilibrium position; k is the constant of spring, k > 0.
According to Newton’s second law,
, (59)
The acceleration is
. (60)
Having inserted (58) and (60) into (59), we receive:
;
. (61)
84
Let us introduce the notation:
2
mk
02 xx
)sin( tAx
)cos( tAx
The interval of time
T = 2π/ω is called the oscillation period, i.e. it is the time
of a full-wave oscillation.
The variable ωt + α is
called the oscillation phase. It
shows the displacement of an
oscillating point at a moment
of time. If t = 0, then
x = Acosα
.
Therefore, equation (61) will have the form:
.
This is the differential equation of harmonic motion. (A moving
system, when described with the help of this equation, is called a simple
harmonic oscillator).
The solution to the equation is a sinusoidal function:
or
These equations are the equations of harmonic oscillations.
Therefore, harmonic oscillations are also called sinusoidal oscillations.
For the sake of simplicity, assume that α = 0. Then, the graph will
have the following view (Fig. 37).
x = Acosωt
t = 0; x = A = max,
where А is the amplitude. This describes the widest displacement of the
oscillating point from the equilibrium position.
If t = 2π/ω, then cosωt = 1 and x = A.
Thus, during the time period t = T = 2π/ω the point returns to the
initial position, and the whole cycle of motion repeats.
85
The quantity α is called the initial phase. It determines the posi-
m
k
k
m
T
2
2
tion of an oscillating point at the initial moment of time.
The angular (circular) frequency is the number of oscillations per
2π/s:
ω = 2π/T.
Alongside the angular frequency ω, it is also possible to introduce the ordinary frequency = 1/T denoting the number of oscillations
per unit of time:
ω = 2πν = 2π/T.
The equation of harmonic motion can be written in an alternative
way:
x = Acos(ωt + α) = Acos(2πνt + α) = Acos(2πt/T + α).
The frequency and the oscillation periods only depend on the dynamic characteristics of the problem – the mass m and the coefficient of
elasticity k:
2
ω
= k / m;
;
.
6.2. Physical and Mathematical Pendulums
It was mentioned earlier that harmonic oscillations are produced
under the action of quasi-elastic forces. Now we will show that forces
which act upon a pendulum, bringing about small angles of deflection,
are quasi-elastic and, as a result, pendulum oscillations are harmonic.
A physical pendulum is an absolutely rigid body which oscillates
under the action of gravitational force about a fixed horizontal axis which
does not cross the center of inertia (gravity).
Fig. 38 demonstrates the notation conventions: O is the axis of
rotation; C is the center of inertia; l is the distance between the axis of
rotation and the center of inertia.
Assume that the pendulum deflected at an angle from the vertical position. The restoring force will include the gravitational component Р1 = mgsin.
86
The motion of the pendulum
mglM
IM
2
2
dt
d
mglI
I
mgl
0
I
mgl
2
Imgl
02
)cos(
max
t
Fig. 38
O C P = mg
P
1
P
2
l
can be considered as a rotational motion of the body about a fixed axis.
The pendulum is acted upon by the
moment of force М = Р1l = mglsin.
For small angles
(sin ), М = mgl.
This moment tends to return
the pendulum into the equilibrium
position and in this respect it is similar to a quasi-elastic force. For this
reason, it is necessary to add a minus
to the moment M and the angular displacement :
. (62)
The equation of gyrodynamics is as follows:
, (63)
viz. I is the moment of inertia; is angular acceleration,
. (64)
Having inserted (62) and (63), we will obtain:
;
After introducing the notation
differential equation of harmonic motion:
Its solution is has the following expression:
where
is the maximum (amplitude) deviation of the pendulum from
max
the equilibrium position.
,
.
, we will receive the
,
87
The oscillation period of the physical pendulum is:
l
2
2
mg
I
T
LmlI
g
L
T 2
2
mlI
g
l
T
2
2
0
. (65)
We will denote
, then
,
specifically, L is the equivalent length of a compound pendulum, equal to
the length of a simple pendulum, whose oscillation period coincides with
the oscillation period of the physical pendulum.
The mathematical pendulum is a point particle, suspended by a
weightless inextensible cord, performing oscillation in the vertical plane
under the action of gravitational force.
In practice, the mathematical pendulum can be regarded as a
body suspended by a light cord whose length by far exceeds the size of
the body. The mathematical pendulum can be considered as a particular
example of the physical pendulum. The moment of inertia for the mathematical pendulum is:
. (66)
Having inserted formula (66) into (65) for the oscillation period
of the mathematical pendulum, we will have:
.
The oscillation period of the mathematical pendulum depends on
its own length and on gravitational acceleration at the particular place of
the Earth. The period T depends neither on mass nor on the amplitude.
The measurements of an oscillation period can be used for the g
determination. These measurements are absolutely precise, for this reason
even slight g fluctuations can be detected. These measurements provide
the basis for determining techniques for identifying the Earth’s configurations and gravimetric prospecting of mineral deposits. Changes in Gvalues, by far exceeding the accuracy tolerances of measurement techniques, can occur due to the presence of more or less compact rocks under the Earth’s surface.
88
6.3. Velocity, Acceleration and Energy of Harmonic Oscillations
)cos( tAx
)sin( tAx
dt
dx
v
xtAx
dt
dx
a
22
)cos(
tAx cos
)
2
cos(sin
tAtAv
)cos(cos
22
tAxtAa
t
x
v
a
/2
3/2
2
Fig. 39
During harmonic motion, the displacement of an oscillating point
from the equilibrium position has the following expression:
.
Then, the velocity and acceleration can be expressed through the
derivatives of displacement with respect to time:
; (67)
. (68)
It follows from equations (67) and (69) that the velocity and acceleration of the point, being in harmonic oscillating motion, are periodic
functions of time with the same frequency as displacement x.
Displacement x, velocity v and acceleration a are phase-shifted in
relation to each other (Fig. 39, 40).
Assume that the initial phase is = 0, then
;
Acceleration is always
proportional and is directed in opposition to displacement (to the equilibrium position). Now we will see how the energy of harmonic oscillating
motion is expressed.
89
Being in the oscillating motion, the point particle possesses velocity
2
2
ê
mv
W
)(sin
2
1
222
tmA
kxf
kxdxdxfdA
2
2
0
ï
kx
kxdxWA
x
)(cos
2
1
)(cos
2
1
2
22222
2
tmAtkA
kx
W
n
x = 0
v = max
a = 0
a
v
x = max
v = 0
a = max
Fig. 40
Fig. 41
t
/2 3/2
2 W Wk
Wp
and, as a result, it has kinetic energy:
.
As the point particle has a different velocity at different positions, it means that its kinetic energy changes with time. It is obvious
that under these conditions the kinetic energy turns into potential energy. If there is no energy loss (the oscillations are sustained), the total
energy remains constant.
The potential energy is measured by the work of external forces
which is performed in order to bring about a particular displacement x:
The total energy is
;
;
;
.
90
êï
WWW
22
2
1
mA
2
2
2
2
2
2
1
A
T
mkA
,
viz. Wk and Wp are phase-shifted by π/2 (Fig. 41); Wk and Wp change
at the frequency of 2ω, which is twice higher than the frequency of the
harmonic oscillation.
6.4. Summation of Oscillations with Equal Direction and Frequency
Let us examine a geometric method of oscillation representation
with the help of an amplitude vector (Fig. 42).
From the point O, we will draw a vector at angle, which is numerically equal to the amplitude. We will rotate this vector at an angular
velocity counterclockwise. At a moment t the angle becomes equal to
t + , and the vector projection А will be
Х = Аcos(t + ). (64)
Consequently, the amplitude vector projection onto the axis x is a
geometric equivalent of the harmonic oscillations described with equation
(69).
It is often necessary to deal with such motion when the body is
involved in two or several oscillations simultaneously (for example, when
a load is suspended by a spring to the ceiling of a wagon with leaf spring
suspensions).
Let us analyze which resultant motion occurs as a result of oscillation summation. We will consider two oscillations equal in direction and
frequency, but having a phase shift and different amplitudes:
x1 = А1cos(t + 1); x2 = А2cos(t + 2)
The displacement x involving the body in both oscillations will
be expressed through the algebraic sum of displacements x1 and x2:
x = x1 + x2 = А1cos(t + 1) + А2cos(t + 2)
We will show this summation in graphic form (Fig. 43).
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