Oscillation physics formulas. Oscillations and waves, laws and formulas. Radius of Newton's dark rings in reflected light

Harmonic vibrations occur according to the law:

x = A cos (ω t + φ 0),

where x- displacement of a particle from the equilibrium position, A- vibration amplitude, ω - angular frequency, φ 0 - initial phase, t- time.

Oscillation period T = .

Oscillating particle speed:

υ = = – Aω sin (ω t + φ 0),

acceleration a = = –Aω 2 cos (ω t + φ 0).

Kinetic energy of a particle making oscillatory motion: E k = =
sin 2 (ω t+ φ 0).

Potential energy:

E n =
cos 2 (ω t + φ 0).

Pendulum oscillation periods

- spring T =
,

where m- the mass of the cargo, k- coefficient of spring stiffness,

- mathematical T = ,

where l- suspension length, g- acceleration of gravity,

- physical T =
,

where I- the moment of inertia of the pendulum relative to the axis passing through the suspension point, m Is the mass of the pendulum, l- the distance from the suspension point to the center of mass.

The reduced length of a physical pendulum is found from the condition: l np = ,

the designations are the same as for the physical pendulum.

When two harmonic oscillations of the same frequency and one direction are added, a harmonic oscillation of the same frequency with an amplitude is obtained:

A = A 1 2 + A 2 2 + 2A 1 A 2 cos (φ 2 - φ 1)

and the initial phase: φ = arctan
.

where A 1 , A 2 - amplitudes, φ 1, φ 2 - the initial phases of the added oscillations.

The trajectory of the resulting movement when adding mutually perpendicular oscillations of the same frequency:

+ – cos (φ 2 - φ 1) = sin 2 (φ 2 - φ 1).

Damped oscillations occur according to the law:

x = A 0 e - β t cos (ω t + φ 0),

where β is the damping coefficient, the meaning of the remaining parameters is the same as for harmonic oscillations, A 0 - initial amplitude. At a moment in time t vibration amplitude:

A = A 0 e - β t .

The damping logarithmic decrement is called:

λ = ln
= β T,

where T- oscillation period: T = .

The quality factor of an oscillatory system is called:

The equation of a plane traveling wave has the form:

y = y 0 cos ω ( t ± ),

where at- displacement of the fluctuating quantity from the equilibrium position, at 0 - amplitude, ω - angular frequency, t- time, NS Is the coordinate along which the wave propagates, υ - wave propagation speed.

The "+" sign corresponds to a wave propagating against the axis X, the sign "-" corresponds to a wave propagating along the axis NS.

The wavelength is called its spatial period:

λ = υ T,

where υ - the speed of wave propagation, T–Period of propagating oscillations.

The wave equation can be written:

y = y 0 cos 2π (+).

A standing wave is described by the equation:

y = (2y 0 cos ) cos ω t.

The amplitude of the standing wave is enclosed in brackets. The points with the maximum amplitude are called antinodes,

x n = n ,

points with zero amplitude - nodes,

x y = ( n + ) .

Examples of problem solving

Assignment 20

The amplitude of the harmonic vibrations is 50 mm, the period is 4 s and the initial phase ... a) Write down the equation of this oscillation; b) find the displacement of the oscillating point from the equilibrium position at t= 0 and for t= 1.5 s; c) draw a graph of this movement.

Solution

The oscillation equation is written as x = a cos ( t+  0).

By condition, the oscillation period is known. Through it, you can express the circular frequency  = . The rest of the parameters are known:

a) x= 0.05 cos ( t + ).

b) Displacement x at t= 0.

x 1 = 0.05 cos = 0.05 = 0.0355 m.

At t= 1.5 s

x 2 = 0.05 cos ( 1,5 + ) = 0.05 cos  = - 0.05 m.

v ) function graph x= 0.05cos ( t + ) as follows:

Let's define the position of several points. Known NS 1 (0) and NS 2 (1.5), as well as the oscillation period. Hence, through  t= 4 s value NS repeats, and after  t = 2 c changes sign. Between the high and low in the middle is 0.

Assignment 21

The point makes a harmonic vibration. The oscillation period is 2 s, the amplitude is 50 mm, the initial phase is zero. Find the speed of a point at the time when its displacement from the equilibrium position is 25 mm.

Solution

1 way. We write the equation of the oscillation of the point:

x= 0.05 cos  t, because  = =.

Find the speed at the moment in time t:

υ = = – 0,05 cos  t.

We find the moment in time when the displacement is 0.025 m:

0.025 = 0.05 cos  t 1 ,

hence cos  t 1 = ,  t 1 = . Substitute this value into the expression for speed:

υ = - 0.05  sin = - 0,05  = 0.136 m / s.

Method 2. Total energy of vibrational motion:

E =
,

where a- amplitude,  - circular frequency, m – particle mass.

At each moment of time, it is the sum of the potential and kinetic energy of the point

E k = , E n = , but k = m 2, then E n =
.

Let's write the law of conservation of energy:

= +
,

from here we get: a 2  2 = υ 2 +  2 x 2 ,

υ = 
= 
= 0.136 m / s.

Assignment 22

Amplitude of harmonic vibrations of a material point A= 2 cm, total energy E= 3 ∙ 10 -7 J. At what displacement from the equilibrium position the force acts on the oscillating point F = 2.25 ∙ 10 -5 N?

Solution

The total energy of a point performing harmonic oscillations is equal to: E =
. (13)

The modulus of elastic force is expressed through the displacement of points from the equilibrium position x in the following way:

F = k x (14)

Formula (13) includes the mass m and the angular frequency , and in (14) - the stiffness coefficient k... But the circular frequency is related to m and k:

 2 = ,

from here k = m 2 and F = m 2 x... By expressing m 2 from relation (13) we obtain: m 2 = , F = x.

From where we get the expression for the displacement x: x = .

Substitution numerical values gives:

x =
= 1.5 ∙ 10 -2 m = 1.5 cm.

Assignment 23

The point participates in two oscillations with the same periods and initial phases. Oscillation amplitudes A 1 = 3 cm and A 2 = 4 cm. Find the amplitude of the resulting oscillation if: 1) oscillations occur in one direction; 2) vibrations are mutually perpendicular.

Solution

If oscillations occur in one direction, then the amplitude of the resulting oscillation will be determined as:

where A 1 and A 2 - amplitudes of the added oscillations,  1 and  2 - initial phases. By condition, the initial phases are the same, which means  2 -  1 = 0, and cos 0 = 1.

Hence:

A =
=
= A 1 +A 2 = 7 cm.

If the vibrations are mutually perpendicular, then the equation of the resulting motion will be:

cos ( 2 -  1) = sin 2 ( 2 -  1).

Since by the condition  2 -  1 = 0, cos 0 = 1, sin 0 = 0, the equation will be written in the form:
=0,

or
=0,

or
.

The resulting relationship between x and at can be plotted on a graph. It can be seen from the graph that the resulting oscillation of a point on a straight line MN... The amplitude of this fluctuation will be defined as: A =
= 5 cm.

Assignment 24

Damped oscillation period T= 4 s, logarithmic damping decrement  = 1.6, initial phase is zero. Point offset at t = equals 4.5 cm. 1) Write the equation of this oscillation; 2) Build a graph of this movement for two periods.

Solution

The equation of damped oscillations with a zero initial phase has the form:

x = A 0 e -  t cos2 .

There are not enough initial amplitude values ​​to substitute numerical values A 0 and damping coefficient .

The damping factor can be determined from the ratio for the logarithmic damping decrement:

 = T.

Thus  = = = 0.4 s -1.

The initial amplitude can be determined by substituting the second condition:

4.5cm = A 0
cos 2 = A 0
cos = A 0
.

From here we find:

A 0 = 4,5∙

(cm) = 7.75 cm.

The final equation of motion is:

x = 0,0775
cost.

Assignment 25

What is the logarithmic damping decrement mathematical pendulum if for t = 1 min, the amplitude of the oscillations decreased by half? Pendulum length l = 1 m.

Solution

The logarithmic damping decrement can be found from the relation:  =  T,

where  is the attenuation coefficient, T- period of fluctuations. Natural circular frequency of a mathematical pendulum:

 0 =
= 3.13 s -1.

The damping coefficient of oscillations can be determined from the condition: A 0 = A 0 e -  t ,

t= ln2 = 0.693,

 =
= 0.0116c -1.

Since <<  0 , то в формуле  =
can be neglected in comparison with  0 and the oscillation period can be determined by the formula: T = = 2c.

Substitute  and T into the expression for the logarithmic damping decrement and we get:

 = T= 0.0116 s -1 ∙ 2 s = 0.0232.

Assignment 26

The equation of sustained oscillations is given in the form x= 4 sin600  t cm.

Find the displacement from the equilibrium position of a point located at a distance l= 75 cm from the source of vibration, after t= 0.01 s after the beginning of the oscillation. Vibration propagation speed υ = 300 m / s.

Solution

Let us write the equation of the wave propagating from a given source: x= 0.04 sin 600  ( t– ).

We find the phase of the wave at a given time in a given place:

t– = 0,01 –= 0,0075 ,

600 ∙ 0.0075 = 4.5,

sin 4,5 = sin = 1.

Therefore, the offset of the point x= 0.04 m, i.e. on distance l = 75 cm from the source at the time t= 0.01 s maximum point shift.

Bibliography

Volkenstein V.S.... Collection of problems for the general course of physics. - SPb .: SpetsLit, 2001.

Saveliev I.V... Collection of questions and problems in general physics. - M .: Nauka, 1998.

Harmonic Equation

where NS - displacement of the oscillating point from the equilibrium position;
t- time; A,ω, φ- respectively amplitude, angular frequency,
initial phase of oscillations; - phase of oscillations at the moment t.

Angular vibration frequency

where ν and T are the frequency and period of the oscillations.

The speed of a point making harmonic oscillations

Harmonic Acceleration

Amplitude A the resulting oscillation, obtained by adding two oscillations with the same frequencies, occurring along one straight line, is determined by the formula

where a 1 and A 2 - the amplitudes of the vibration components; φ 1 and φ 2 are their initial phases.

The initial phase φ of the resulting oscillation can be found from the formula

The frequency of beats arising from the addition of two oscillations occurring along one straight line with different, but close in value, frequencies ν 1 and ν 2,

The equation of the trajectory of a point participating in two mutually perpendicular oscillations with amplitudes A 1 and A 2 and initial phases φ 1 and φ 2,

If the initial phases φ 1 and φ 2 of the vibration components are the same, then the trajectory equation takes the form

that is, the point moves in a straight line.

In the event that the phase difference, the equation
takes the form

that is, the point moves along an ellipse.

Differential equation of harmonic vibrations of a material point

Or ,
where m is the mass of the point; k - quasi-elastic force coefficient ( k=Tω 2).

The total energy of a material point performing harmonic vibrations,

The period of oscillation of a body suspended on a spring (spring pendulum),

where m- body mass; k - spring rate. The formula is valid for elastic vibrations within the limits in which Hooke's law is fulfilled (with a small mass of the spring in comparison with the mass of the body).

The period of oscillation of a mathematical pendulum

where l- the length of the pendulum; g - acceleration of gravity. The period of oscillation of a physical pendulum

where J- moment of inertia of the oscillating body about the axis

fluctuations; a- the distance of the center of mass of the pendulum from the axis of oscillation;

Reduced length of a physical pendulum.

The formulas given are exact for the case of infinitesimal amplitudes. At finite amplitudes, these formulas give only approximate results. At amplitudes no more than the error in the value of the period does not exceed 1%.

The period of torsional vibrations of a body suspended on an elastic thread,

where J - moment of inertia of the body about the axis coinciding with the elastic thread; k - the stiffness of the elastic thread, equal to the ratio of the elastic moment that occurs when the thread is twisted to the angle through which the thread is twisted.

Differential Equation of Damped Oscillations
, or ,

where r- coefficient of resistance; δ - attenuation coefficient:; ω 0 - natural angular frequency of oscillations *

Damped oscillation equation

where A (t) - amplitude of damped oscillations at the moment t;ω is their angular frequency.

Angular frequency of damped oscillations

О Dependence of the amplitude of damped oscillations on time

where A 0 - vibration amplitude at the moment t=0.

Logarithmic decrement of fluctuations

where A (t) and A (t + T) - the amplitudes of two successive oscillations spaced in time from each other by a period.

Forced oscillation differential equation

where is the external periodic force acting on
fluctuating material point and causing forced
fluctuations; F 0 - its amplitude value;

Forced vibration amplitude

Resonant frequency and resonant amplitude and

Examples of problem solving

Example 1. The point oscillates according to the law x (t) =, where A = 2 see Determine the initial phase φ if

x(0) = cm and NS , (0)<0. Построить векторную диаграмму для мо-­
cop t=0.

Solution. Let's use the equation of motion and express the displacement at the moment t= 0 through the initial phase:

From here we find the initial phase:

* In the earlier formulas of harmonic vibrations, the same value was denoted simply by ω (without index 0).

Substitute the given values ​​into this expression x(0) and A:φ=
=. The value of the argument is satisfied
two angle values:

In order to decide which of these values ​​of the angle φ satisfies
also raises the condition, we first find:

Substituting in this expression the value t= 0 and alternately values
initial phases and, we find

As always A> 0 and ω> 0, then the condition is satisfied only
to the first value of the initial phase.
Thus, the required initial
phase

Based on the found value of φ,
them a vector diagram (Fig. 6.1).
Example 2. Material point
mass T= 5 g performs a harmonic
vibrations with frequency ν = 0.5 Hz.
Amplitude of vibration A= 3 cm. Op-
Determine: 1) the speed υ point in the
time moment when offset x =
= 1.5 cm; 2) maximum strength
F max acting on the point; 3)
Rice. 6.1 full energy E fluctuating point
Ki.

and we obtain the formula for the speed by taking the first time derivative of the displacement:

To express the speed in terms of displacement, it is necessary to exclude time from formulas (1) and (2). To do this, we square both equations, divide the first by A 2, the second on A 2 ω 2 and add:

Or

Solving the last equation for υ , find

Performing calculations using this formula, we get

The plus sign corresponds to the case when the direction of the velocity coincides with the positive direction of the axis NS, minus sign - when the direction of speed coincides with the negative direction of the axis NS.

The displacement at harmonic vibration, in addition to equation (1), can also be determined by the equation

Repeating the same solution with this equation, we get the same answer.

2. The force acting on a point is found according to Newton's second law:

where a - point acceleration, which we obtain by taking the time derivative of the velocity:

Substituting the expression for acceleration into formula (3), we obtain

Hence the maximum value of the force

Substituting into this equation the values ​​of the quantities π, ν, T and A, find

3. The total energy of an oscillating point is the sum of the kinetic and potential energies calculated for any moment of time.

It is easiest to calculate the total energy at the moment when the kinetic energy reaches its maximum value. At this moment, the potential energy is zero. Therefore, the total energy E oscillating point is equal to the maximum kinetic energy

The maximum speed is determined from formula (2) by setting
:. Substituting the expression for speed in the form
mule (4), find

Substituting the values ​​of the quantities into this formula and making calculations, we get

or μJ.

Example 3. l= 1 m and mass m 3 = 400 g small balls fortified with masses m 1 = 200 gi m 2 = 300g. The rod vibrates about the horizontal axis, perpendicular

dicular to the rod and passing through its middle (point O in Fig. 6.2). Determine the period T vibrations made by the rod.

Solution. The oscillation period of a physical pendulum, which is a rod with balls, is determined by the ratio

Where J - T - its mass; l С - distance from the center of mass of the pendulum to the axis.

The moment of inertia of a given pendulum is equal to the sum of the moments of inertia of the balls J 1 and J 2 and rod J 3:

Taking the balls as material points, we express the moments of their inertia:

Since the axis passes through the middle of the bar, then
its moment of inertia about this axis J 3 =
= . Substituting the resulting expressions J 1 , J 2 and
J 3 into formula (2), we find the total moment of inertia of the
physical pendulum:

Making calculations using this formula, we find

Rice. 6.2 The mass of the pendulum consists of the masses of the balls and the mass
rod:

Distance l C We find the center of mass of the pendulum from the axis of oscillations based on the following considerations. If the axis NS direct along the bar and align the origin with the point O, then the required distance l is equal to the coordinate of the center of mass of the pendulum, i.e.

Substituting the values ​​of the quantities m 1 , m 2 , m, l and making calculations, we find

Making calculations according to formula (1), we obtain the oscillation period of the physical pendulum:

Example 4. The physical pendulum is a rod
the length l= 1 m and mass 3 T 1 with attached to one of its ends
hoop with diameter and weight T 1 . Horizontal axis Oz

the pendulum passes through the middle of the rod perpendicular to it (Fig. 6.3). Determine the period T oscillations of such a pendulum.

Solution. The oscillation period of a physical pendulum is determined by the formula

(1)

where J - the moment of inertia of the pendulum relative to the oscillation axis; T - its mass; l C - distance from the center of mass of the pendulum to the axis of oscillation.

The moment of inertia of the pendulum is equal to the sum of the moments of inertia of the rod J 1 and hoop J 2:

The moment of inertia of the rod about the axis,
perpendicular to the bar and passing
through its center of mass, is determined by the form-
le. In this case t = 3T 1 and

We find the moment of inertia of the hoop, use
called Steiner's theorem,
where J - moment of inertia relative to
arbitrary axis; J 0 - moment of inertia
with respect to the axis passing through the center of mass
parallel to a given axis; a - distance
between the specified axes. Applying this form-
mule to the hoop, we get

Rice. 6.3

Substituting expressions J 1 and J 2 into formula (2), we find the moment of inertia of the pendulum relative to the axis of rotation:

Distance l C from the axis of the pendulum to its center of mass is

Substituting into formula (1) expressions J, lс and the mass of the pendulum, we find the period of its oscillations:

After calculating by this formula, we get T= 2.17 s.

Example 5. Two vibrations of the same direction are added
niya expressed by equations; x 2 =
=, where A 1 = 1cm, A 2 = 2 cm, s, s, ω =
=. 1. Determine the initial phases φ 1 and φ 2 of the components of the

baths. 2. Find the amplitude A and the initial phase φ of the resulting wobble. Write the equation for the resulting fluctuation.

Solution. 1. The equation of harmonic oscillation has the form

We transform the equations given in the problem statement to the same form:

Comparing expressions (2) with equality (1), we find the initial phases of the first and second oscillations:

Glad and glad.

2. To determine the amplitude A of the resulting fluctuation, it is convenient to use the vector diagram presented on rice. 6.4. According to the cosine theorem, we get

where is the phase difference of the components of the oscillations.
Since, then, substituting the found
the values ​​of φ 2 and φ 1 will be rad.

Rice. 6.4

Substitute the values A 1 , A 2 and into formula (3) and
let's make calculations:

A = 2.65 cm.

The tangent of the initial phase φ of the resulting oscillation determines
lim directly from Fig. 6.4: , otku-
yes initial phase

Substitute the values A 1 , A 2 , φ 1, φ 2 and make calculations:

Since the angular frequencies of the added vibrations are the same,
then the resulting vibration will have the same frequency ω. it
allows you to write the equation of the resulting oscillation in the form
, where A= 2.65 cm,, glad.

Example 6. The material point participates simultaneously in two mutually perpendicular harmonic oscillations, the equations of which

where a 1 = 1 cm, A 2 = 2 cm,. Find the equation of the trajectory of the point
Ki. Draw trajectory to scale and specify
direction of movement of the point.

Solution. To find the equation of the trajectory of a point, we exclude the time t from the given equations (1) and (2). To do this, use

we study the formula. In this case
, therefore

Since according to formula (1) , then the trajectory equation
ri

The resulting expression is the equation of a parabola, the axis of which coincides with the axis Oh. From equations (1) and (2) it follows that the displacement of a point along the coordinate axes is limited and is in the range from -1 to +1 cm along the axis Oh and from -2 to +2 cm along the axis OU.

To construct the trajectory, we find by equation (3) the values y, corresponding to a series of values NS, satisfying the condition, see, and draw up a table:

In order to indicate the direction of movement of a point, follow how its position changes over time. At the initial moment t= 0 point coordinates are equal x(0) = 1 cm and y(0) = 2 cm.At the next moment in time, for example, at t 1 = l s, the coordinates of the points will change and become equal NS(1) = -1 cm, y ( t )=0. Knowing the position of the points at the initial and subsequent (close) moments of time, you can indicate the direction of movement of the point along the trajectory. In fig. 6.5 this direction of movement is indicated by an arrow (from the point A to the origin). After at the moment t 2 = 2 s the oscillating point reaches the point D, it will move in the opposite direction.

Kinematics of harmonic vibrations

6.1. The equation of vibrations of a point has the form,
where ω = π s -1, τ = 0.2 s. Determine the period T and the initial phase φ
hesitation.

6.2. Determine the period T, frequency v and initial phase φ of oscillations, given by the equation, where ω = 2.5π s -1,
τ = 0.4 s.

6.3.
where A x (0) = 2 mass media
; 2) x (0) = cm and; 3) x (0) = 2cm and; 4)
x (0) = u. Construct vector diagram for
moment t=0.

6.4. The point vibrates. According to the law,
where A= 4 cm. Determine the initial phase φ if: 1) x (0) = 2 mass media
; 2) x(0) = cm and; 3) NS(0) = cm and;
4) x(0) = cm and. Construct vector diagram for
moment t=0.

6.5. The point vibrates according to the law,
where A= 2 cm; ; φ = π / 4 rad. Build dependency graphs
from time: 1) displacement x (t); 2) speed; 3) acceleration

6.6. The point oscillates with an amplitude A= 4 cm and period T = 2 s. Write the equation of these vibrations, assuming that in
moment t= 0 offset x (0) = 0 and . Determine the phase
for two points in time: 1) when the displacement x = 1cm and;
2) when the speed = -6 cm / s and x<0.

6.7. The point moves uniformly around the circle counterclockwise with a period of T = 6 s. Diameter d circle is 20 cm.Write the equation of motion of the projection of a point on the axis NS, passing through the center of the circle, if at the moment of time taken as the initial, the projection onto the axis NS is zero. Find offset NS, the speed and acceleration of the projection of the point at the moment t = 1c.

6.8. Determine the maximum values ​​of the speed and acceleration of a point performing harmonic oscillations with amplitude A = 3cm and corner frequency

6.9. The point oscillates according to the law, where A =
= 5 cm; ... Determine the acceleration of a point at a point in time,
when its speed = 8 cm / s.

6.10. The point performs harmonic oscillations. The greatest
bias x m ax points is 10 cm, the highest speed =
= 20 cm / s. Find the angular frequency ω of oscillations and the maximum acceleration of the point.

6.11. The maximum speed of a point performing harmonic oscillations is 10 cm / s, the maximum acceleration =
= 100 cm / s 2. Find the angular frequency ω of oscillations, their period T
and amplitude A. Write the equation of oscillations, taking the initial phase equal to zero.

6.12. The point oscillates according to the law. At some point in time, the offset NS 1 point turned out to be equal to 5 cm.When the phase of the oscillations doubled, the displacement x became equal to 8 cm.Find the amplitude A hesitation.

6.13. The point oscillates according to the law.
At some point in time, the offset NS point is 5 cm, its speed
= 20 cm / s and acceleration = -80 cm / s 2. Find the amplitude A, angular frequency ω, period T oscillations and phase at the considered moment of time.

Addition of vibrations

6.14. Two identically directed harmonic oscillations of the same period with amplitudes A 1 = 10 cm and A 2 = 6 cm add up to one vibration with an amplitude A = 14 cm. Find the phase difference of the added oscillations.

6.15. Two harmonic vibrations, directed along one straight line and having the same amplitudes and periods, add up to one vibration of the same amplitude. Find the phase difference of the added oscillations.

6.16. Determine the amplitude A and the initial phase φ of the result
oscillating oscillation arising from the addition of two oscillations
the same direction and period: and
, where A 1 =A 2 = 1 cm; ω = π s -1; τ = 0.5 s. Find the equation of the resulting oscillation.

6.17. The point participates in two equally directed oscillations: and, where a 1 = 1cm; A 2 = 2 cm; ω =
= 1 s -1. Determine the amplitude A the resulting fluctuation,
its frequency v and initial phase φ. Find the equation of this motion.

6.18. Two harmonic vibrations add up, one per
reigns with the same periods T 1 =T 2 = 1.5 s and amplitudes
A 1 = A 2 = 2cm. The initial phases of oscillations and. Determine the amplitude A and the initial phase φ of the resulting wobble. Find its equation and plot it to scale
vector diagram of the addition of amplitudes.

6.19. Three harmonic oscillations of the same direction with the same periods are added T 1 = T 2 = T 3 = 2 s and amplitudes A 1 =A 2 =A 3 = 3 cm. The initial phases of oscillations are φ 1 = 0, φ 2 = π / 3, φ 3 = 2π / 3. Build a vector diagram of the addition of amplitudes. Determine the amplitude from the drawing A and the initial phase φ of the resulting wobble. Find his equation.

6.20. Add two harmonic vibrations of the same
frequency and the same direction: and x 2 =
=. Draw a vector diagram for a moment
time t= 0. Determine analytically the amplitude A and initial
phase φ of the resulting oscillation. Postpone A and φ on the vector
diagram. Find the equation of the resulting oscillation (in trigonometric form through the cosine). Solve the problem for two
cases: 1) A 1 = 1cm, φ 1 = π / 3; A 2 = 2 cm, φ 2 = 5π / 6; 2) A 1 = 1cm,
φ 1 = 2π / 3; A 2 = 1 cm, φ 2 = 7π / 6.

6.21. Two tuning forks sound simultaneously. The frequencies ν 1 and ν 2 of their oscillations are respectively equal to 440 and 440.5 Hz. Determine the period T beats.

6.22. Two mutually perpendicular vibrations add up,
expressed by equations and, where
a 1 =2 cm, A 2 = 1 cm, τ = 0.5 s. Find the equation of the trajectory
and build it, showing the direction of movement of the point.

6.23. The point performs simultaneously two harmonic oscillations occurring in mutually perpendicular directions
and expressed by equations and,
where a 1 = 4 cm, A 1 = 8 cm,, τ = 1 s. Find the equation of the trajectory of a point and build a graph of its movement.

6.24. The point performs simultaneously two harmonic oscillations of the same frequency, occurring in mutually perpendicular directions expressed by the equations: 1) and

Find (for eight cases) the equation of the trajectory of the point, build it with respect to the scale and indicate the direction of movement. Accept: A = 2 cm, A 1 = 3 cm, A 2 = 1cm; φ 1 = π / 2, φ 2 = π.

6.25 ... The point participates simultaneously in two mutually perpendicular oscillations, expressed by the equations and
, where A 1 = 2 cm, A 2 = 1 cm.Find the trajectory equation
point and build it, indicating the direction of movement.

6.26. A point simultaneously performs two harmonic oscillations occurring in mutually perpendicular directions
and expressed by equations and, where A 1 =
= 0.5 cm; A 2 = 2 cm.Find the equation of the trajectory of the point and construct
her, indicating the direction of movement.

6.27. The motion of a point is given by the equations and y =
=, where A 1 = 10 cm, A 2 = 5 cm, ω = 2 s -1, τ = π / 4 s. Find
the equation of the trajectory and velocity of a point at the moment of time t= 0.5 s.

6.28. The material point participates simultaneously in two mutually perpendicular oscillations, expressed by the equations
and where A 1 =2 cm, A 2 = 1 cm. Find
equation of the trajectory and build it.

6.29. The point participates simultaneously in two harmonic oscillations occurring in mutually perpendicular directions described by the equations: 1) and

Find the equation of the trajectory of a point, build it with respect to the scale and indicate the direction of movement. Accept: A= 2 cm; A 1 = s cm.

6.30. The point participates simultaneously in two mutually perpendicular
oscillations expressed by equations and

y = A 2 sin 0.5ω t, where A 1 = 2cm, A 2 = 3 cm. Find the equation of the trajectory of the point and construct it, indicating the direction of motion.

6.31. The shift of the luminous point on the oscilloscope screen is the result of the addition of two mutually perpendicular oscillations, which are described by the equations: 1) x = A sin 3 ω t and at=A sin 2ω t; 2) x = A sin 3ω t and y=A cos 2ω t; 3) x = A sin 3ω t and y = A cos ω t.

Using the graphical addition method and observing the scale, construct the trajectory of the luminous point on the screen. Accept A= 4 cm.

Dynamics of harmonic vibrations. Pendulums

6.32. Material point by mass T= 50 g performs oscillations, the equation of which has the form x = A cos ω t, where A= 10 cm, ω = 5 s -1. Find strength F, acting on the point, in two cases: 1) at the moment when the phase ω t= π / 3; 2) at the position of the greatest point displacement.

6.33. Oscillations of a material point with mass T= 0.1 g occur according to the equation NS=A cos ω t, where A= 5 cm; ω = 20 s -1. Determine the maximum values ​​of the restoring force F max and kinetic energy T m ah.

6.34. Find the restoring force F in the moment t= 1 s and full energy E material point oscillating according to the law x = A cos ω t, where A = 20 cm; ω = 2π / 3 s -1. Weight T material point is equal to 10 g.

6.35. Oscillations of a material point occur according to the equation x = A cos ω t, where A= 8 cm, ω = π / 6 s -1. The moment when the restoring force F for the first time reached a value of -5 mN, the potential energy of the P point became equal to 100 μJ. Find this moment in time t and the corresponding phase ω t.

6.36. Weight weight m= 250 g, suspended from a spring, oscillates vertically with a period T = 1with. Determine the stiffness k springs.

6.37. A weight was suspended from the coil spring, as a result of which the spring was stretched by x = 9 see what will be the period T oscillation of the weight, if it is pulled down a little and then released?

6.38. A weight suspended from a spring vibrates vertically with an amplitude A= 4 cm. Determine the total energy E oscillations of the weight, if the stiffness k the spring is 1 kN / m.

6.39. Find the ratio of the lengths of two mathematical pendulums if the ratio of the periods of their oscillations is 1.5.

6.40. l = 1m installed in the elevator. The elevator rises with acceleration a= 2.5 m / s 2. Determine the period T oscillations of the pendulum.

6.41. At the ends of a thin rod of length l= 30 cm, identical weights are attached, one at each end. A rod with weights vibrates about a horizontal axis passing through a point d = 10 cm from one of the ends of the rod. Determine the reduced length L and period T oscillations of such a physical pendulum. Disregard the mass of the rod.

6.42. On a rod long l= 30 cm two identical weights are fixed: one - in the middle of the rod, the other - at one of its ends. A rod with a weight oscillates about a horizontal axis passing through the free end of the rod. Determine the reduced length L and period T vibrations of such a system. Disregard the mass of the rod.

6.43. A system of three weights connected by rods of length l= 30 cm (Fig. 6.6), oscillates about the horizontal axis passing through point O perpendicular to the plane of the drawing. Find period T system fluctuations. We neglect the masses of the rods, consider the weights as material points.

6.44. A thin hoop, hung on a nail, driven horizontally into the wall, oscillates in a plane parallel to the wall. Radius R the hoop is 30 cm.Calculate the period T hoop vibrations.

Rice. 6.6

Rice. 6.7

6.45. Homogeneous disc with radius R= 30 cm oscillates about a horizontal axis passing through one of the generatrices of the cylindrical surface of the disk. What is the period T his hesitation?

6.46. Disc radius R = 24cm vibrates about a horizontal axis passing through the middle of one of the radii perpendicular to the plane of the disk. Determine the reduced length L and period T oscillations of such a pendulum.

6.47. From a thin homogeneous disk with a radius R= 20 cm cut out a part that looks like a circle with a radius r = 10cm, as shown in fig. 6.7. The rest of the disc oscillates about the horizontal axis O, which coincides with one of the generatrices of the cylindrical surface of the disc. Find period T oscillations of such a pendulum.

6.48. Mathematical pendulum length l 1 = 40 cm and a physical pendulum in the form of a thin straight rod long l 2 = 60 cm oscillate synchronously about the same horizontal axis. Determine the distance a the center of mass of the rod from the axis of vibration.

6.49. Physical pendulum in the form of a thin straight rod with a length l= 120 cm oscillates about a horizontal axis passing perpendicular to the rod through a point some distance away a from the center of mass of the rod. At what value a period T fluctuation has the least value?

6.50. T with a small ball of mass fixed on it T. The pendulum oscillates about a horizontal axis passing through point O on the rod. Determine the period T harmonic oscillations of the pendulum for cases a, b, c, d shown in Fig. 6.8. Length l the rod is equal to 1 m. The ball is considered as a material point.

Rice. 6.9

Rice. 6.8

6.51. A physical pendulum is a thin homogeneous rod with a mass T with two small balls fixed on it T and 2 T... The pendulum oscillates about a horizontal axis passing through a point O on the rod. Determine the frequency ν of harmonic oscillations of the pendulum for the cases a B C D, shown in Fig. 6.9. Length l the rod is equal to 1 m. The balls are considered as material points.

6.52. Body mass T= 4 kg, fixed on the horizontal axis, oscillated with a period T 1 = 0.8 s. When a disk was mounted on this axis so that its axis coincided with the axis of vibration of the body, the period T 2 oscillations became equal to 1.2 s. Radius R disk is equal to 20 cm, its mass is equal to the mass of the body. Find the moment of inertia J body relative to the axis of vibration.

6.53. Mass hydrometer T= 50 g, having a tube of diameter d= 1 cm, floats in water. The hydrometer was slightly immersed in water and then left to itself, as a result of which it began to perform harmonic oscillations. Find period T these fluctuations.

6.54. In a U-tube open at both ends with a cross-sectional area S= 0.4 cm 2, mercury is quickly poured in a mass T= 200 g. Determine the period T fluctuations of mercury in the tube.

6.55. The swollen log, the cross-section of which is constant along its entire length, plunged vertically into the water so that only a small part of it (in comparison with the length) is above the water. Period T vibration of the log is 5 s. Determine the length l logs.

Damped Oscillations

6.56. The amplitude of the damped oscillations of the pendulum during the time t 1= 5 minutes decreased by half. In what time t 2, counting from the initial moment, will the amplitude decrease by eight times?

6.57. During t= 8 min, the amplitude of the damped oscillations of the pendulum decreased three times. Determine the attenuation coefficient δ .

6.58. Amplitude of oscillations of a pendulum length l = 1m per time t= 10 minutes decreased by half. Determine the logarithmic decrement of fluctuations Θ.

6.59. The logarithmic decrement of oscillations Θ of the pendulum is 0.003. Determine the number N full oscillations, which the pendulum must make so that the amplitude is halved.

6.60. Kettlebell mass T= 500 g suspended from a coil spring with a stiffness k= 20 N / m and performs elastic vibrations in a certain medium. Logarithmic decrement of fluctuations Θ = 0.004. Determine the number N total vibrations that the weight must perform in order for the vibration amplitude to decrease by n= 2 times. How long does it take t will this decrease occur?

6.61. Body mass T= 5 g performs damped oscillations. For a time t = 50s the body has lost 60% of its energy. Determine the coefficient of resistance b.

6.62. Determine the period T damped oscillations, if the period T 0 natural oscillations of the system is equal to 1 s and the logarithmic decrement of oscillations Θ = 0.628.

6.64. Body mass T= 1 kg is in a viscous medium with a drag coefficient b= 0.05 kg / s. With two identical springs with a stiffness k= 50 N / m, each body is held in equilibrium, while the springs are not deformed (Fig. 6.10). The body was displaced from the equilibrium position and

released. Determine: 1) damping coefficient δ ; 2) the frequency ν of oscillations; 3) logarithmic decrement of fluctuations Θ; 4) number N oscillations, after which the amplitude will decrease by a factor of e.

Forced vibrations. Resonance

6.65. Under the action of the gravity of the electric motor, the cantilever beam on which it is installed bent over h= 1 mm. At what speed NS the motor armature can there be a resonance hazard?

6.66. Wagon weight T= 80 t has four springs. Rigidity k the springs of each spring are equal to 500 kN / m. At what speed will the wagon begin to swing strongly due to jolts at the rail joints, if the length l rail is 12.8 m?

6.67. The oscillating system performs damped oscillations with a frequency of ν = 1000 Hz. Determine the frequency ν 0 of natural vibrations if the resonant frequency ν pe s = 998 Hz.

6.68. Determine how much the resonant frequency differs from the frequency ν 0 = l kHz of natural oscillations of the system, characterized by the damping coefficient δ = 400 s -1.

6.69. Determine the logarithmic decrement of oscillations Θ of the oscillatory system, for which resonance is observed at a frequency lower than the natural frequency ν 0 = 10 kHz by Δν = 2 Hz.

6.70. Period T 0 of the natural oscillations of the spring pendulum is 0.55 s. In a viscous environment, the period T the same pendulum became equal to 0.56 s. Determine the resonant frequency ν pe of oscillations.

6.71. Spring pendulum (rigidity k spring is 10 N / m, weight T load is equal to 100 g) makes forced vibrations in a viscous medium with a drag coefficient r= 2 · 10 -2 kg / s. Determine the damping factor δ and the resonant amplitude A res, if the amplitude value of the driving force F 0 = 10 mN.

6.72. The body makes forced vibrations in a medium with a drag coefficient r = 1g / s. Considering the damping to be small, determine the amplitude value of the driving force if the resonance amplitude A res = 0.5 cm and the frequency ν 0 of natural vibrations is 10 Hz.

6.73. The amplitudes of forced harmonic oscillations at a frequency of ν 1 = 400 Hz and ν 2 = 600 Hz are equal to each other. Determine the resonant frequency ν pe s. Disregard attenuation.

6.74. To a coil spring with a stiffness k = 10N / m suspended a weight T= 10 g and immersed the entire system in a viscous medium. Adopting a coefficient of resistance b equal to 0.1 kg / s, determine: 1) the frequency ν 0 of natural vibrations; 2) resonant frequency ν pe s; 3) resonant amplitude A cut, if the driving force changes according to the harmonic law and its amplitude value F 0 == 0.02 N; 4) the ratio of the resonant amplitude to the static displacement under the action of the force F 0.

6.75. How many times will the amplitude of the forced oscillations be less than the resonant amplitude if the frequency of the change in the driving force is greater than the resonant frequency: 1) by 10%? 2) twice? The damping coefficient δ in both cases is taken equal to 0.1 ω 0 (ω 0 is the angular frequency of natural oscillations).

Harmonic vibrations - vibrations performed according to the laws of sine and cosine. The following figure shows a graph of the change in the coordinate of a point over time according to the cosine law.

picture

Amplitude of vibration

The amplitude of the harmonic vibration is the greatest value of the displacement of the body from the equilibrium position. The amplitude can take on different values. It will depend on how much we displace the body at the initial moment of time from the equilibrium position.

The amplitude is determined by the initial conditions, that is, the energy imparted to the body at the initial moment of time. Since sine and cosine can take values ​​in the range from -1 to 1, then the equation must have a factor Xm, which expresses the amplitude of the oscillations. Equation of motion for harmonic vibrations:

x = Xm * cos (ω0 * t).

Oscillation period

The oscillation period is the time of one complete oscillation. The oscillation period is denoted by the letter T. The units of the period correspond to the units of time. That is, in SI, these are seconds.

Oscillation frequency - the number of oscillations made per unit of time. The vibration frequency is denoted by the letter ν. The oscillation frequency can be expressed in terms of the oscillation period.

ν = 1 / T.

Frequency units in SI 1 / sec. This unit of measurement is called Hertz. The number of oscillations in a time of 2 * pi seconds will be equal to:

ω0 = 2 * pi * ν = 2 * pi / T.

Oscillation frequency

This value is called the cyclic vibration frequency. In some literature, the name circular frequency is found. The natural frequency of an oscillating system is the frequency of free oscillations.

The natural frequency is calculated by the formula:

The natural frequency depends on the properties of the material and the mass of the load. The higher the stiffness of the spring, the higher the natural frequency. The greater the mass of the load, the lower the frequency of natural vibrations.

These two conclusions are clear. The stiffer the spring, the more acceleration it will impart to the body when the system is unbalanced. The greater the mass of the body, the slower this speed of this body will change.

Free oscillation period:

T = 2 * pi / ω0 = 2 * pi * √ (m / k)

It is noteworthy that at small angles of deflection, the period of oscillation of the body on the spring and the period of oscillation of the pendulum will not depend on the amplitude of the oscillations.

Let us write down the formulas for the period and frequency of free oscillations for a mathematical pendulum.

then the period will be

T = 2 * pi * √ (l / g).

This formula will be valid only for small deflection angles. From the formula we see that the oscillation period increases with the length of the pendulum thread. The longer the length, the slower the body will oscillate.

The period of oscillations does not depend at all on the mass of the load. But it depends on the acceleration of gravity. As g decreases, the oscillation period will increase. This property is widely used in practice. For example, to measure the exact value of free acceleration.

Period.

Period T is called the time interval during which the system makes one complete oscillation:

N- the number of complete oscillations during t.

Frequency.

Frequency ν is the number of oscillations per unit of time:

Frequency unit - 1 hertz (Hz) = 1 s -1

Cyclic frequency:

Harmonic oscillation equation:

x- displacement of the body from position. X m is the amplitude, that is, the maximum displacement, (ω t+ φ 0) is the oscillation phase, Ψ 0 is its initial phase.

Speed.

For φ 0 = 0:

Acceleration.

For φ 0 = 0:

Free vibrations.

Free vibrations are those that arise in a mechanical system (oscillator) with a unit deviation from the equilibrium position, having a natural frequency ω 0, set only by the parameters of the system, and damping over time due to the presence of friction.

Mathematical pendulum.

Frequency:

l- the length of the pendulum, g- acceleration of gravity.

The pendulum has the maximum kinetic energy at the moment of passing the equilibrium position:

Spring pendulum.

Frequency:

k- spring stiffness, m- the mass of the cargo.

The pendulum has the maximum potential energy at the maximum displacement:

Forced vibrations.

Forced vibrations are those that arise in an oscillatory system (oscillator) under the influence of a periodically changing external force.

Resonance.

Resonance - a sharp increase in amplitude X m of forced vibrations when the frequency ω of the driving force coincides with the frequency ω 0 of the natural vibrations of the system.

Waves.

Waves are vibrations of matter (mechanical) or fields (electromagnetic) that propagate in space over time.

Wave speed.

The speed of propagation of the wave υ is the speed of transmission of the vibration energy. In this case, the particles of the medium vibrate about the equilibrium position, and do not move with the wave.

Wavelength.

Wavelength λ is the distance over which the oscillation propagates in one period:

The unit of wavelength is 1 meter (m).

Wave frequency:

The unit of wave frequency is 1 hertz (Hz).

(lat. amplitude- value) is the greatest deviation of the oscillating body from the equilibrium position.

For a pendulum, this is the maximum distance the ball moves from its equilibrium position (figure below). For oscillations with small amplitudes, such a distance can be taken as the length of the arc 01 or 02, and the lengths of these segments.

The amplitude of oscillations is measured in units of length - meters, centimeters, etc. On the oscillation graph, the amplitude is defined as the maximum (modulo) ordinate of a sinusoidal curve (see figure below).

Period of hesitation.

Oscillation period- this is the smallest period of time after which the system, performing oscillations, again returns to the same state in which it was at the initial moment of time, chosen arbitrarily.

In other words, the oscillation period ( T) Is the time during which one complete oscillation is completed. For example, in the figure below, this is the time during which the weight of the pendulum moves from the extreme right point through the equilibrium point O to the leftmost point and back through the point O back to the far right.

Thus, for a full period of oscillation, the body travels a path equal to four amplitudes. The oscillation period is measured in units of time - seconds, minutes, etc. The oscillation period can be determined from the well-known oscillation graph (see figure below).

Strictly speaking, the concept of "oscillation period" is valid only when the values ​​of the oscillating quantity are exactly repeated after a certain period of time, that is, for harmonic oscillations. However, this concept also applies to cases of approximately repeating quantities, for example, for damped oscillations.

Oscillation frequency.

Oscillation frequency Is the number of vibrations per unit of time, for example, 1 s.

The SI unit of frequency is called hertz(Hz) in honor of the German physicist G. Hertz (1857-1894). If the vibration frequency ( v) is equal to 1 Hz, then this means that one oscillation is performed for every second. The frequency and period of oscillations are related by the relations:

In the theory of vibrations, they also use the concept cyclical, or circular frequency ω ... It is related to the usual frequency v and the oscillation period T ratios:

.

Cyclic frequency Is the number of oscillations made during 2π seconds.