Oscillations and waves. Equation of vibrations Mechanical vibrations and waves

As you study this section, please keep in mind that fluctuations of different physical nature are described from common mathematical positions. Here it is necessary to clearly understand such concepts as harmonic oscillation, phase, phase difference, amplitude, frequency, oscillation period.

It must be borne in mind that in any real oscillatory system there is resistance of the medium, i.e. the oscillations will be damped. To characterize the damping of oscillations, a damping coefficient and a logarithmic damping decrement are introduced.

If oscillations occur under the influence of an external, periodically changing force, then such oscillations are called forced. They will be undamped. The amplitude of forced oscillations depends on the frequency of the driving force. As the frequency of forced oscillations approaches the frequency of natural oscillations, the amplitude of forced oscillations increases sharply. This phenomenon is called resonance.

When moving on to the study of electromagnetic waves, you need to clearly understand thatelectromagnetic waveis an electromagnetic field propagating in space. The simplest system emitting electromagnetic waves is an electric dipole. If a dipole undergoes harmonic oscillations, then it emits a monochromatic wave.

Formula table: oscillations and waves

Physical laws, formulas, variables

Oscillation and wave formulas

Harmonic vibration equation:

where x is the displacement (deviation) of the fluctuating quantity from the equilibrium position;

A - amplitude;

ω - circular (cyclic) frequency;

α - initial phase;

(ωt+α) - phase.

Relationship between period and circular frequency:

Frequency:

Relationship between circular frequency and frequency:

Periods of natural oscillations

1) spring pendulum:

where k is the spring stiffness;

2) mathematical pendulum:

where l is the length of the pendulum,

g - free fall acceleration;

3) oscillatory circuit:

where L is the inductance of the circuit,

C is the capacitance of the capacitor.

Natural frequency:

Addition of oscillations of the same frequency and direction:

1) amplitude of the resulting oscillation

where A 1 and A 2 are the amplitudes of the vibration components,

α 1 and α 2 - initial phases of the vibration components;

2) the initial phase of the resulting oscillation

Equation of damped oscillations:

e = 2.71... - the base of natural logarithms.

Amplitude of damped oscillations:

where A 0 is the amplitude at the initial moment of time;

β - attenuation coefficient;

Attenuation coefficient:

oscillating body

where r is the resistance coefficient of the medium,

m - body weight;

oscillatory circuit

where R is active resistance,

L is the inductance of the circuit.

Frequency of damped oscillations ω:

Period of damped oscillations T:

Logarithmic damping decrement:

Harmonic Equation

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

Angular frequency

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

The speed of a point performing harmonic oscillations is

Acceleration during harmonic oscillation

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 - amplitudes of 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 that arise when adding two oscillations occurring along one straight line with different but similar frequencies ν 1 and ν 2,

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 oscillation 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 is , the equation
takes the form

that is, the point moves along an ellipse.

Differential equation of harmonic oscillations 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 oscillations is

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

Where m- body mass; k- spring stiffness. The formula is valid for elastic vibrations within the limits in which Hooke's law is satisfied (with a small mass of the spring compared to the mass of the body).

Period of oscillation of a mathematical pendulum

Where l- length of the pendulum; g- acceleration of gravity. Period of oscillation of a physical pendulum

Where J- moment of inertia of the oscillating body relative to the axis

hesitation; A- distance of the center of mass of the pendulum from the axis of oscillation;

Reduced length of a physical pendulum.

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

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

Where J- moment of inertia of the body relative to the axis coinciding with the elastic thread; k- the rigidity of an elastic thread, equal to the ratio of the elastic moment arising when the thread is twisted to the angle at which the thread is twisted.

Differential equation of damped oscillations
, or ,

Where r- resistance coefficient; 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 - amplitude of oscillations at moment t=0.

Logarithmic oscillation decrement

Where A(t) And A (t+T) - amplitudes of two successive oscillations separated in time by a period.

Differential equation of forced oscillations

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

Amplitude of forced oscillations

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 X , (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 previously given formulas for harmonic vibrations, the same quantity was designated simply ω (without the index 0).

Let's substitute the given values into this expression x(0) and A:φ=
= . The value of the argument satisfies
two angle values:

In order to decide which of these angle values φ satisfies
also satisfies the condition, let’s find first:

Substituting the value into this expression t=0 and alternate values
initial phases and , we find

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

Based on the found value of φ, we construct
them a vector diagram (Fig. 6.1).
Example 2. Material point
mass T=5 g performs harmonic
oscillations with frequency ν =0.5 Hz.
Oscillation amplitude A=3 cm. Op-
divide: 1) speed υpoints in mo-
moment in time when the offset x=
= 1.5 cm; 2) maximum strength
F max acting on a point; 3)
Rice. 6.1 total energy E oscillating point
ki.

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

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

Having solved the last equation for υ , we'll find

Having performed 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 X, minus sign - when the direction of velocity coincides with the negative direction of the axis X.

The displacement during harmonic oscillation, 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. We find the force acting on a point using Newton’s second law:

Where A - acceleration of the point, which we obtain by taking the time derivative of the speed:

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

Hence the maximum value of force

Substituting the values of π, ν into this equation, T And A, we'll find

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

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

We determine the maximum speed from formula (2), putting
: . Substituting the expression for speed in the form
mulu (4), let's find

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

or µJ.

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

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

Solution. The period of oscillation of a physical pendulum, such as a rod with balls, is determined by the relation

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

The moment of inertia of this pendulum is equal to the sum of the moments of inertia of the balls J 1 and J2 and rod J 3:

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

Since the axis passes through the middle of the rod, then
its moment of inertia about this axis J 3 =
= . Substituting the resulting expressions J 1 , J2 And
J 3 into formula (2), we find the total moment of inertia of the fi-
static pendulum:

Having carried out 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 will find the center of mass of the pendulum from the axis of oscillation based on the following considerations. If the axis X direct along the rod and align the origin of coordinates with the point ABOUT, then the required distance l 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 after performing calculations, we find

Having made calculations using formula (1), we obtain the oscillation period of a physical pendulum:

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

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

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

(1)

Where J- moment of inertia of the pendulum relative to the axis of oscillation; T - its mass; l C - the 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:

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

We will find the moment of inertia of the hoop using
called Steiner's theorem,
Where J- moment of inertia relative to the pro-
arbitrary axis; J 0 - moment of inertia relative to
relative to the axis passing through the center of mass
parallel to a given axis; A - distance
between the indicated axes. Using 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 equal to

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

After calculating using this formula we get T=2.17 s.

Example 5. Two oscillations of the same direction are added
tions 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 oscillatory components

Baniya. 2. Find the amplitude A and the initial phase φ of the resulting oscillation. Write the equation for the resulting vibration.

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

Let us transform the equations specified in the problem statement to the same form:

From a comparison of 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 oscillation, it is convenient to use the vector diagram presented in rice. 6.4. According to the cosine theorem, we get

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

Rice. 6.4

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

A= 2.65 cm.

The tangent of the initial phase φ of the resulting oscillation is determined
lim directly from fig. 6.4: , from
yes initial phase

Let's substitute the values A 1 , A 2 , φ 1, φ 2 and perform the calculations:

Since the angular frequencies of the added oscillations are the same,
then the resulting oscillation will have the same frequency ω. This
allows us to write the equation of the resulting vibration in the form
, Where A=2.65 cm, , rad.

Example 6. A 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. Construct a trajectory respecting the scale and indicate
direction of movement of the point.

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

Let's use the formula. In this case
, That's why

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

The resulting expression is the equation of a parabola whose axis 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 ranges from -1 to +1 cm along the axis Oh and from -2 to +2 cm along the axis OU.

To construct the trajectory, we use equation (3) to find the values y, corresponding to a range of values X, satisfying the condition cm, and create a table:

In order to indicate the direction of movement of a point, we will monitor 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 a subsequent point in time, for example when t 1 =l s, the coordinates of the points will change and become equal X(1)= -1 cm, y( t )=0. Knowing the positions 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 the moment t 2 = 2 s the oscillating point will reach the point D, it will move in the opposite direction.

Kinematics of harmonic oscillations

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

6.2. Define 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)= and . Construct a vector diagram for
moment t=0.

6.4. The point oscillates 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) X(0)= cm and ;
4) x(0)= cm and . Construct a vector diagram for
moment t=0.

6.5. The point oscillates 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 an equation for these oscillations, assuming that in
moment t=0 offset x(0)=0 And . Determine phase
for two moments in time: 1) when the displacement x= 1cm and ;
2) when 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 onto the axis X, passing through the center of the circle, if at the moment of time taken as the initial, the projection onto the axis X equal to zero. Find offset X, speed and acceleration of the projection of a point at an instant t= 1s.

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

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

6.10. The point performs harmonic oscillations. Greatest
bias x m ah point is 10 cm, maximum 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, 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 displacement X 1 point turned out to be equal to 5 cm. When the oscillation phase 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 displacement X point is 5 cm, its speed
= 20 cm/s and acceleration = -80 cm/s 2. Find amplitude A, angular frequency ω, period T oscillations and phase at the considered moment in 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 amplitude A= 14 cm. Find the phase difference of the added oscillations.

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

6.16. Determine amplitude A and the initial phase f results
oscillating vibration that occurs when two vibrations are added
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 vibration.

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

6.18. Two harmonic oscillations of one are added
reigns with equal periods T 1 =T 2 =1.5 s and amplitudes
A 1 =A 2 = 2cm. Initial phases of oscillations and. Determine amplitude A and the initial phase φ of the resulting oscillation. Find its equation and construct it to scale
vector diagram of amplitude addition.

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

6.20. Two harmonic oscillations of the same
frequency and same direction: and x 2 =
= . Draw a vector diagram for the 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 vibration (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 440 and 440.5 Hz. Define period T beats.

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

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

6.24. A point simultaneously performs 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 a point, construct it in accordance with 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
points and construct it, indicating the direction of movement.

6.26. A point simultaneously performs two harmonic oscillations occurring in mutually perpendicular directions
and expressed by the 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
equation of trajectory and speed of a point at an instant of time t=0.5 s.

6.28. A material point participates simultaneously in two mutually perpendicular oscillations, expressed by the equations
and where A 1 =2 cm, A 2 =1 cm. Find
trajectory equation and construct 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 the point, construct it in accordance with the scale and indicate the direction of movement. Accept: A=2 cm; A 1 =z cm.

6.30. The point participates simultaneously in two mutually perpendicular
cular vibrations 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 movement.

6.31. The displacement 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 method of addition and observing the scale, construct the trajectory of a luminous point on the screen. Accept A=4 cm.

Dynamics of harmonic oscillations. Pendulums

6.32. Material point with mass T=50 g undergoes 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 a point, in two cases: 1) at the moment when the phase ω t=π/3; 2) in the position of the greatest displacement of the point.

6.33. Oscillations of a material point with mass T=0.1 g occur according to the equation X=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 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. At the moment when the restoring force F for the first time reached a value of -5 mN, the potential energy P of the point became equal to 100 μJ. Find this moment in time t and its corresponding phase ω t.

6.36. A weight weighing m=250 g, suspended from a spring, oscillates vertically with a period T= 1With. Determine hardness k springs.

6.37. A weight was suspended from a spiral spring, causing the spring to stretch by x=9 see What will be the period T Does the weight oscillate if you pull it down a little and then release it?

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

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

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

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

6.42. On a rod length 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 reduced length L and period T vibrations of such a system. Neglect 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 a horizontal axis passing through point O perpendicular to the drawing plane. Find period T system vibrations. Neglect the masses of the rods; consider the loads 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 vibrations of the hoop.

Rice. 6.6

Rice. 6.7

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

6.47. From a thin homogeneous disk with a radius R= 20 cm, a part is cut out that looks like a circle with a radius r= 10cm, as shown in Fig. 6.7. The remaining part of the disk oscillates relative to the horizontal axis O, which coincides with one of the generatrices of the cylindrical surface of the disk. 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 of length l 2 =60 cm oscillate synchronously about the same horizontal axis. Determine distance A the center of mass of the rod from the axis of vibration.

6.49. A physical pendulum in the form of a thin straight rod of 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 oscillations has the least significance?

6.50. T with a small ball of mass attached to it T. The pendulum oscillates about a horizontal axis passing through point O on the rod. Define period T harmonic oscillations of the pendulum for cases a, b, c, d shown in Fig. 6.8. Length l the length of the rod is 1 m. Consider the ball 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 mass balls attached to it T and 2 T. The pendulum oscillates about a horizontal axis passing through a point ABOUT 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 length of the rod is 1 m. Consider the balls as material points.

6.52. Body mass T=4 kg, fixed on a 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 the disk is 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 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 vibrations. Find period T these fluctuations.

6.54. Into a U-shaped tube, open at both ends, with a cross-sectional area S=0.4 cm 2 quickly pour in a mass of mercury T=200 g. Determine the period T vibrations of mercury in the tube.

6.55. A swollen log, the cross-section of which is constant along its entire length, is immersed vertically in water so that only a small part (compared to its length) is above the water. Period T vibration of the log is 5 s. Determine length l logs

Damped oscillations

6.56. Amplitude of damped oscillations of a pendulum over time t 1=5 min decreased by half. In what time t2, counting from the initial moment, the amplitude will decrease by eight times?

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

6.58. The amplitude of oscillations of a pendulum with a length l= 1m per time t=10 min decreased by half. Determine the logarithmic decrement of oscillations Θ.

6.59. The logarithmic decrement of oscillations Θ of the pendulum is 0.003. Determine the number N total oscillations that the pendulum must make in order for the amplitude to be halved.

6.60. Weight mass T=500 g suspended from a coil spring with stiffness k=20 N/m and performs elastic vibrations in a certain medium. Logarithmic decrement of oscillations Θ=0.004. Determine the number N full oscillations that the weight must make in order for the amplitude of oscillations to decrease by n=2 times. For how long 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 drag coefficient b.

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

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

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

Forced vibrations. Resonance

6.65. Under the influence of gravity of the electric motor, the cantilever beam on which it is installed bent by h=1 mm. At what speed P Is there a danger of resonance in the motor armature?

6.66. Car weight T=80 t has four springs. Rigidity k springs of each spring is 500 kN/m. At what speed will the car begin to sway strongly due to shocks at the rail joints, if the length l rail is 12.8 m?

6.67. The oscillatory system performs damped oscillations with a frequency of ν=1000 Hz. Determine the frequency ν 0 of natural oscillations if the resonant frequency ν pe з =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 a 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 natural oscillations of a spring pendulum are equal to 0.55 s. In a viscous medium, the period T of the same pendulum became equal to 0.56 s. Determine the resonant frequency ν pe of oscillations.

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

6.72. A body performs forced oscillations in a medium with a drag coefficient r= 1g/s. Assuming the attenuation to be small, determine the amplitude value of the driving force if the resonant amplitude A res =0.5 cm and the frequency ν 0 of natural oscillations is 10 Hz.

6.73. The amplitudes of forced harmonic oscillations at frequencies ν 1 =400 Hz and ν 2 =600 Hz are equal. Determine the resonant frequency ν pe h. Neglect damping.

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

6.75. How many times will the amplitude of forced oscillations be less than the resonant amplitude if the frequency of change of 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 oscillations are oscillations performed according to the laws of sine and cosine. The following figure shows a graph of changes in the coordinates of a point over time according to the cosine law.

picture

Oscillation amplitude

The amplitude of a harmonic vibration is the greatest value of the displacement of a body from its 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, the equation must contain a factor Xm, expressing the amplitude of the oscillations. Equation of motion for harmonic vibrations:

x = Xm*cos(ω0*t).

Oscillation period

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

Oscillation frequency is the number of oscillations performed per unit of time. The oscillation frequency is designated by the letter ν. The oscillation frequency can be expressed in terms of the oscillation period.

ν = 1/T.

Frequency units are 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 quantity is called the cyclic frequency of oscillations. In some literature the name circular frequency appears. The natural frequency of an oscillatory system is the frequency of free oscillations.

The frequency of natural oscillations is calculated using the formula:

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

These two conclusions are obvious. The stiffer the spring, the greater the acceleration it will impart to the body when the system is thrown out of balance. The greater the mass of a body, the slower the 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's write down the formulas for the period and frequency of free oscillations for a mathematical pendulum.

then the period will be equal

T = 2*pi*√(l/g).

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

The period of oscillation does not depend at all on the mass of the load. But it depends on the acceleration of free fall. 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.

Harmonic oscillations occur according to the law:

x = A cos(ω t + φ 0),

Where x– displacement of the particle from the equilibrium position, A– amplitude of oscillations, ω – circular frequency, φ 0 – initial phase, t- time.

Oscillation period T = .

Speed of oscillating particle:

υ = = – Aω sin(ω t + φ 0),

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

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

Potential energy:

E n=
cos 2 (ω t + φ 0).

Periods of pendulum oscillations

– spring T =
,

Where m– mass of cargo, k– spring stiffness coefficient,

– mathematical T = ,

Where l– suspension length, g- acceleration of gravity,

– physical T =
,

Where I– moment of inertia of the pendulum relative to the axis passing through the suspension point, m– mass of the pendulum, l– 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 a physical pendulum.

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

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

and initial phase: φ = arctan
.

Where A 1 , A 2 – amplitudes, φ 1, φ 2 – initial phases of folded 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 logarithmic damping decrement is called:

λ = log
= 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 oscillating quantity from the equilibrium position, at 0 – amplitude, ω – angular frequency, t- time, X– coordinate along which the wave propagates, υ – speed of wave propagation.

The “+” sign corresponds to a wave propagating against the axis X, the “–” sign corresponds to a wave propagating along the axis X.

The wavelength is called its spatial period:

λ = υ T,

Where υ – wave propagation speed, 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 0cos ) cos ω t.

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

x n = n ,

points with zero amplitude - nodes,

x y = ( n + ) .

Examples of problem solving

Problem 20

The amplitude of harmonic oscillations 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 at t= 1.5 s; c) draw a graph of this movement.

Solution

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

According to the condition, the period of oscillation is known. Through it we can express the circular frequency  = . The remaining parameters are known:

A) x= 0.05cos( t + ).

b) Offset 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 ) graph of a function x=0.05cos ( t + ) as follows:

Let's determine the position of several points. Known X 1 (0) and X 2 (1.5), as well as the oscillation period. So, through  t= 4 s value X repeats, and after  t = 2 s changes sign. Between the maximum and minimum in the middle is 0.

Problem 21

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

Solution

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

x= 0.05 cos t, because  = =.

Finding the speed at the moment of 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 = . We substitute this value into the expression for speed:

υ = – 0.05  sin = – 0.05  = 0.136 m/s.

Method 2. Total energy of oscillatory motion:

E =
,

Where A– amplitude,  – circular frequency, m – particle mass.

At each moment of time it consists of the potential and kinetic energy of the point

E k = , E n = , But k = m 2, which means E n =
.

Let's write down the law of conservation of energy:

= +
,

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

υ = 
= 
= 0.136 m/s.

Problem 22

Amplitude of harmonic oscillations of a material point A= 2 cm, total energy E= 3∙10 -7 J. At what displacement from the equilibrium position does the force act 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 mass m and circular 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. Having expressed m 2 from relation (13) we obtain: m 2 = , F = x.

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

Substituting the numeric values gives:

x =
= 1.5∙10 -2 m = 1.5 cm.

Problem 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 vibration if: 1) the vibrations occur in one direction; 2) the vibrations are mutually perpendicular.

Solution

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

Where A 1 and A 2 – amplitudes of added oscillations,  1 and  2 – initial phases. According to the 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 oscillations are mutually perpendicular, then the equation of the resulting motion will be:

cos( 2 –  1) = sin 2 ( 2 –  1).

Since by condition  2 –  1 = 0, cos 0 = 1, sin 0 = 0, the equation will be written as:
=0,

or
=0,

or
.

The resulting relationship between x And at can be depicted on a graph. The graph shows that the result will be a oscillation of a point on a straight line MN. The amplitude of this oscillation is determined as: A =
= 5 cm.

Problem 24

Period of damped oscillations T=4 s, logarithmic damping decrement  = 1.6, initial phase is zero. Point displacement at t = equals 4.5 cm. 1) Write the equation of this vibration; 2) Construct a graph of this movement for two periods.

Solution

The equation of damped oscillations with 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 attenuation coefficient .

The attenuation coefficient can be determined from the relation for the logarithmic attenuation decrement:

 = T.

Thus  = = = 0.4 s -1 .

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

4.5 cm = 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.

Problem 25

What is the logarithmic damping decrement of a mathematical pendulum, if for t = 1 min amplitude of 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 oscillation. Natural circular frequency of a mathematical pendulum:

 0 =
= 3.13 s -1 .

The oscillation damping coefficient 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 compared to  0 and the oscillation period can be determined by the formula: T = = 2c.

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

 = T= 0.0116 s -1 ∙ 2 s = 0.0232.

Problem 26

The equation of undamped 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 vibration source, through t= 0.01 s after the start of oscillation. Oscillation propagation speed υ = 300 m/s.

Solution

Let us write down the equation of a 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 point displacement x= 0.04 m, i.e. on distance l =75 cm from the source at time t= 0.01 s maximum point displacement.

Bibliography

Volkenshtein V.S.. Collection of problems for the general course of physics. – St. Petersburg: SpetsLit, 2001.

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

Harmonic oscillations occur according to the law:

x = A cos(ω t + φ 0),

Where x– displacement of the particle from the equilibrium position, A– amplitude of oscillations, ω – circular frequency, φ 0 – initial phase, t- time.

Oscillation period T = .

Speed of oscillating particle:

υ = = – Aω sin(ω t + φ 0),

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

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

Potential energy:

E n=
cos 2 (ω t + φ 0).

Periods of pendulum oscillations

– spring T =
,

Where m– mass of cargo, k– spring stiffness coefficient,

– mathematical T = ,

Where l– suspension length, g- acceleration of gravity,

– physical T =
,

Where I– moment of inertia of the pendulum relative to the axis passing through the suspension point, m– mass of the pendulum, l– 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 a physical pendulum.

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

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

and initial phase: φ = arctan
.

Where A 1 , A 2 – amplitudes, φ 1, φ 2 – initial phases of folded 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 logarithmic damping decrement is called:

λ = log
= 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 ± ),

The “+” sign corresponds to a wave propagating against the axis X, the “–” sign corresponds to a wave propagating along the axis X.

The wavelength is called its spatial period:

λ = υ T,

Where υ – wave propagation speed, 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 0cos ) cos ω t.

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

x n = n ,

points with zero amplitude - nodes,

x y = ( n + ) .

Examples of problem solving

Problem 20

Solution

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

According to the condition, the period of oscillation is known. Through it we can express the circular frequency  = . The remaining parameters are known:

A) x= 0.05cos( t + ).

b) Offset 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 ) graph of a function x=0.05cos ( t + ) as follows:

Problem 21

Solution

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

x= 0.05 cos t, because  = =.

Finding the speed at the moment of 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 = . We substitute this value into the expression for speed:

υ = – 0.05  sin = – 0.05  = 0.136 m/s.

Method 2. Total energy of oscillatory motion:

E =
,

Where A– amplitude,  – circular frequency, m – particle mass.

At each moment of time it consists of the potential and kinetic energy of the point

E k = , E n = , But k = m 2, which means E n =
.

Let's write down the law of conservation of energy:

= +
,

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

υ = 
= 
= 0.136 m/s.

Problem 22

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 mass m and circular 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. Having expressed m 2 from relation (13) we obtain: m 2 = , F = x.

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

Substituting the numeric values gives:

x =
= 1.5∙10 -2 m = 1.5 cm.

Problem 23

Solution

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

Hence:

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

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

cos( 2 –  1) = sin 2 ( 2 –  1).

Since by condition  2 –  1 = 0, cos 0 = 1, sin 0 = 0, the equation will be written as:
=0,

or
=0,

or
.

Problem 24

Solution

The equation of damped oscillations with 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 attenuation coefficient .

The attenuation coefficient can be determined from the relation for the logarithmic attenuation decrement:

 = T.

Thus  = = = 0.4 s -1 .

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

4.5 cm = 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.

Problem 25

What is the logarithmic damping decrement of a mathematical pendulum, if for t = 1 min amplitude of 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 oscillation. Natural circular frequency of a mathematical pendulum:

 0 =
= 3.13 s -1 .