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 weight; 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 a 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:
Or 
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 force value
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 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 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 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 are
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 Oh.
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 cm and
; 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 cm and
; 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 point's trajectory 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 at 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. For how long t2, counting from the initial moment, the amplitude will decrease by eight times?
6.57. During the time 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. Over 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 n 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).
>>Physics: Mechanical vibrations
Oscillation is a very common type of movement. This is the swaying of tree branches in the wind, the vibration of strings of musical instruments, the movement of a piston in a car engine cylinder, the swing of a pendulum in a wall clock, and even the beating of our heart.
Today's lesson topic will be devoted to the study of oscillation and oscillatory movements.
The process of oscillation is the most common type of movement that exists in nature. And if we consider this process from the point of view of mechanical movements, then vibrations can be called the most common type of mechanical movement.
The concept of oscillation is considered to be a movement that is repeated in whole or in part over time.
Do you think oscillatory movements are the swaying of trees or the movement of leaves under the influence of the wind? Naturally, such movement can be attributed to oscillations. Oscillatory movements are also performed by swaying swings, vibrating strings of musical instruments, and the swinging of a pendulum in a clock. And even any movement human body and our heartbeat, which repeats over time, also performs oscillatory movements.
Well, now we can draw a conclusion and define this phenomenon.
A process that repeats itself over time is called oscillation.
Conditions required for oscillation
Now let’s take a closer look at the process of oscillatory movements using the examples of spring and thread pendulums.
Now let's turn our attention to our drawings, which depict these pendulums.
In the first picture we are presented with the so-called thread pendulum; this pendulum is also called a mathematical one. Now let's look at what this mathematical pendulum is. And it represents some kind of massive body, in this case a ball, which is suspended on a long and thin thread. If we try to take it and move it to the side, disturbing its balance, and then release it, then this ball will perform repeated movements to the sides, and at the same time it will periodically pass through the equilibrium position. In this case, we can say that this ball will begin to perform oscillatory movements, that is, oscillate.
Now consider the following figure, which shows a spring pendulum. This pendulum is presented in the form of a weight, which is attached to a spring and, under the action of the elastic force of this spring, is capable of performing oscillatory movements.
But, as you can already see from the examples given, certain conditions are necessary for oscillations to occur.
For oscillations to exist it is necessary:
Firstly, the presence of the oscillatory system itself. And in our case, such a system is these pendulums, which are capable of carrying out these oscillatory movements.
Secondly, it is necessary to have a point of equilibrium and, moreover, a stable equilibrium.
Thirdly, the mandatory presence of energy reserves, with the help of which oscillatory movements will be carried out.
And, fourthly, the presence of a small friction force, since if the friction force is large, then, naturally, there can be no talk of any vibrations.
Units of vibration amplitude
The quantities that characterize oscillatory movements are:
1. Amplitude, which is denoted by the symbol “A” and measured in units of length such as meters, centimeters, etc. As a rule, amplitude is considered to be the maximum distance over which a body oscillates from its equilibrium position.
2. The period, which is designated by the symbol “T” and is measured in time units, that is, in minutes, seconds, etc. The period is the time during which one oscillation occurs.
3.Frequency, which is designated by the symbol “V”. The oscillation frequency is considered to be the number of oscillations that occur in 1 second.
In the SI system, the unit of frequency is usually called the “hertz”. It received its name in honor of the German physicist G. Hertz.
If we assume that the oscillation frequency is equal to 1 Hz, then this will mean that one oscillation occurs in one second. If the frequency is v = 50 Hz, then naturally 50 oscillations will be made per second.
Oscillation amplitude formulas
Now let's move on to considering the vibration formulas. It should be noted here that for the period T and frequency v of oscillations, the same formulas that are used for the period and frequency of rotation will be correct.
Let's consider the meanings of these formulas in more detail:
1. First, in order to find the period of oscillations, we need to take the time t during which a certain number of oscillations were made and divide by n, which is the number of these oscillations, and we get the following formula:
2. Secondly, if we need to find the frequency of oscillations, then we need to take the number of oscillations and divide them by the time during which these oscillations occurred. As a result, we got the following formula:
But to better understand how to count the number of vibrations, you need to have an idea of what one complete vibration is. To do this, let us again return to the consideration of Fig. 30, where we are clearly shown that the pendulum begins its movement from position 1, then it passes through the equilibrium position and moves to position 2, and then it returns from the second position to the equilibrium position and again returns to position 1. This whole process is with one hesitation.
It is worth paying attention to the fact that when comparing these two formulas, the period and frequency of oscillations are mutually inverse quantities, i.e.
Fluctuation chart
As you already know from today's lesson, the position of the body during the process of oscillation is constantly changing.
An oscillation graph is a dependence graph where the coordinates of an oscillating body depend on time.
Now let's look at what a fluctuation graph is. To do this, we will take time t along the horizontal axis of our graph, and place the x coordinate on the vertical axis. Now, using the modulus of this coordinate, we see at what distance from the initial position, that is, the equilibrium position, the oscillating body is located at a given moment in time.
And, when a given body passes through the equilibrium position, then in this case the sign of the coordinate changes to the opposite. That is, this sign shows us that the body has moved to the other side of the equilibrium position.
Practical work
Now let's do some interesting experiments. To do this, let's try to connect a spring pendulum to a writing device. And then we will begin to evenly move the paper tape in front of this oscillating body. If you look closely at Figure 32, you will see how a line appears on the tape using a brush, which will coincide with the oscillation graph.
Figure 33 shows the installation of a string pendulum, where the oscillations of this pendulum can also be recorded. In this example, the pendulum is a funnel filled with sand. In the same way, we place a paper tape under an oscillating funnel and observe how the sand that pours out of the funnel leaves a corresponding mark.
Now we see that over small intervals and with fairly low friction, the oscillation graph of these pendulums is a sinusoid.
So, for example, on the graph we can see all the oscillatory movements, where A = 5 cm, T = 4 s and v = 1/T = 0.25 Hz.
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.
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
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Physical laws, formulas, variables |
Oscillation and wave formulas |
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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. |
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Relationship between period and circular frequency: |
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Frequency: |
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Relationship between circular frequency and frequency: |
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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 circuit inductance, C is the capacitance of the capacitor. |
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Natural frequency: |
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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 |
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Equation of damped oscillations: e = 2.71... - the base of natural logarithms. |
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Amplitude of damped oscillations: where A 0 is the amplitude at the initial moment of time; β - attenuation coefficient; |
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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. |
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Frequency of damped oscillations ω: |
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Period of damped oscillations T: |
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Logarithmic damping decrement: |
(lat. amplitude- magnitude) is the greatest deviation of the oscillating body from the equilibrium position.
For a pendulum, this is the maximum distance that the ball moves away 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, as well as 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 the sinusoidal curve (see figure below).
Oscillation period.
Oscillation period- this is the shortest period of time through which a system oscillating returns again 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 occurs. For example, in the figure below, this is the time it takes for the pendulum bob to move from the rightmost point through the equilibrium point ABOUT to the far left point and back through the point ABOUT again to the far right.
Over a full period of oscillation, the body thus travels a path equal to four amplitudes. The period of oscillation is measured in units of time - seconds, minutes, etc. The period of oscillation can be determined from a well-known graph of oscillations (see figure below).
The concept of “oscillation period”, strictly speaking, is valid only when the values of the oscillating quantity are exactly repeated after a certain period of time, i.e. for harmonic oscillations. However, this concept also applies to cases of approximately repeating quantities, for example, for damped oscillations.
Oscillation frequency.
Oscillation frequency- this is the number of oscillations performed per unit of time, for example, in 1 s.
The SI unit of frequency is named hertz(Hz) in honor of the German physicist G. Hertz (1857-1894). If the oscillation frequency ( v) is equal to 1 Hz, this means that every second there is one oscillation. The frequency and period of oscillations are related by the relations:
In the theory of oscillations they also use the concept cyclical, or circular frequency ω . It is related to the normal frequency v and oscillation period T ratios:
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Cyclic frequency is the number of oscillations performed per 2π seconds