1- 1 The equation of motion of a known particle is. (1) Find the displacement of the particle from t=0 to t= 1. (2) Find the trajectory equation of the particle.

Solution: (1)

The displacement of the particle is

(2) The equation of motion has, and t is eliminated.

The trajectory equation is sum.

The motion equation of 1-2 particle is to find the velocity and acceleration of the particle when t = 0, 1.

Solution: Starting from the definitions of speed and acceleration.

,

So when t=0, the velocity of the particle is; When t= 1, the velocity of the particle is.

The acceleration at both moments is zero.

1-3 The equation of motion of a particle moving on a plane is called [B].

(a) Uniform linear motion (b) Uniform variable speed linear motion

(c) Projectile motion (d) General curvilinear motion

1-4 it is known that a particle moves in a straight line along the Ox axis, and its instantaneous acceleration changes as follows. When t=0, m. Find: (1) the velocity of the particle at time t (2) the motion equation of the particle.

Solution: (1) is derived from

Both sides are integrated at the same time. When the initial condition t=0 is brought into the integral equation, there are

The velocity of a particle at time t is obtained by the following formula

(2) by

Both sides are integrated at the same time. When the initial condition t=0, m is brought into the integral equation, and there are

The equation of motion of particles is obtained as follows

Chapter 4 Mechanical Vibration

4- 1 It is known that there are four particles moving on the X axis, and the relationship between the particle displacement X and the external force F at a certain moment is expressed by the following four equations (where A and B are natural numbers), in which the force that cannot make the particle make a simple harmonic vibration is [C].

(A) (B)

(C) (D)

4-2 In the motions of the following objects, what can be regarded as simple harmonic vibration is [B]

(a) Put the wood block into water, completely immerse it and dive to a certain depth, and then release it.

(b) Place the spring vibrator on a smooth slope and let it vibrate.

(c) Release a small slider from the edge of the smooth semi-circular groove.

(d) the movement of the ball when it is patted.

4-3 For the study of the same simple harmonic vibration, both people chose the equilibrium position as the coordinate origin, but one of them chose the vertical cow axis as the coordinate system, while the other chose the vertical cow axis as the coordinate system, so the different quantities in the vibration equation are [].

Amplitude; (b) cycle frequency;

Initial stage; Amplitude and circumferential frequency.

A: (C)

4-4 An object vibrates simply according to the law of cosine function, and its initial phase is [].

(a) x0 = 0, v0? 0; (B) x0 = 0，v0 & lt0;

(C) x0 = 0，v0 = 0(D) x0 =？ a，v0 = 0 .

A: (A)

4-5 A particle makes a simple harmonic vibration with amplitude a and period t at the initial moment.

(1) The displacement of the particle is A/2, and it moves in the negative direction of the X axis;

(2) The displacement of the particle is -A/2, and it moves along the positive direction of the X axis;

(3) The particle is in an equilibrium position and the velocity is negative;

(4) The particle is in a negative maximum displacement;

Write the simple harmonic dynamic equation and draw the rotation vector diagram when t=0.

Solution: (1) (2)

(3) (4)

4-6 If the particle vibrates simply with the period t, the shortest time for the particle to move forward from the equilibrium position to half the maximum displacement is [C].

(A) (B) (C) (D)

4-7 particles are in simple harmonic vibration, with the same amplitude and period. The vibration equation of the first particle is 0. When the first particle returns to the equilibrium position from a negative displacement relative to its equilibrium position, the second particle is in a positive maximum displacement. Then the vibration equation of the second particle is

(1); (b) and: [] (c); (D) 10 .

Solution: (a) Use the rotation vector method to judge, as shown in the attached figure:

therefore

Answer (1)

4-8 Simple harmonic dynamic curve is shown in the figure. The period of vibration is, and the vibration equation of particle is determined by graph. When t = 2s, the displacement, velocity and acceleration of the particle are.

A:; 0; -0.06 meters? s– 1； 0

4-9 particle vibrates simply along the X axis, with its angular frequency ω = 10 rad/s, its initial displacement x0 = 7.5 cm, and its initial velocity v0 = 75.0cm/s. Try to write the vibration equation of this particle.

Solution: by

Get cm = 0.11m.

According to the initial conditions x0 = 7.5 cm, v0 = 75.0 cm/s, combined with the rotation vector diagram, we can know that,

;

The vibration equation of a particle is m.

4- 10 A particle with a mass of 2 kg vibrates along the X axis according to equation (SI). Find the period, initial phase, maximum speed and maximum acceleration of (1) vibration; (2) The phase and displacement of vibration at t =1s.

Solution: (1) From the vibration equation, the period s of vibration.

From the initial stage of the vibration equation

What's the speed? s- 1

What is the maximum speed? s- 1

The acceleration is m? s-2

Maximum acceleration m? s-2

(2) When t = 1s, the phase of vibration is

The displacement is x = 0.02m

4- 1 1 A particle is in simple harmonic vibration, and the vibration equation is cm. At t (unit: s), at cm, move in the negative direction of X axis. Q: What's the shortest time to get back to this position?

The solution can be obtained by rotating vector method, and the phase at time t is

When I came back again,

The minimum angle of vector rotation is

Minimum time of use, i.e.

So there is

4- 12 A particle with a mass of 0.0 1 kg oscillates simply, with an amplitude of 0. 1m and a maximum kinetic energy of 0.02 j. If the particle is in a negative maximum displacement at the beginning, the vibration equation of the particle is obtained.

Solution: simple harmonic vibration energy conservation, there are

Radians per second

According to the rotation vector diagram:

Therefore, the particle vibration equation is

The fifth chapter mechanical wave

5- 1 Which of the following equations and the movements described in the words is simple harmonic motion? [C ]

(1)

(2)

(c) The waveform is always a plane wave with a sine or cosine curve.

(d) The wave source is a plane wave that vibrates harmoniously, but its amplitude is always attenuated.

The expression of 5-2 plane harmonic is (SI), its angular frequency.

=, wave velocity u =, wavelength? = 。

Solution:? = 125 rad; ，u = 338

17.0m

5-3 When X is a certain value, the physical meaning reflected by the wave equation is [C]

(a) the waveform at a certain moment; The spread of energy.

(c) indicates the vibration law of the particle at x (d) indicates the distribution of the vibration state of each particle.

5-4 It is known that a wave source is located at x = 5 m, and its vibration equation is: (m). When the plane harmonic generated by this wave source propagates forward along the X axis with the wave velocity U, its wave equation is [D].

(A) (B)

(C) (D)

5-5 Wave with frequency of 500Hz, speed of 350m/s, and the distance between two points with phase difference of 2π/3 is _.

Solution:? , = 0.233m.

5-6 A plane harmonic propagates in the negative direction of the X axis. It is known that the vibration equation of a particle at x=- 1m is (SI), and if the wave velocity is u, the expression of this wave is.

A:

5-7 A series of plane harmonics propagate along the X axis without attenuation. The amplitude of the wave is 2× 10-3 m, the period is 0.0 1 s, and the wave speed is 400 m? S- 1. When t = 0, the prime element of the origin of the X axis moves to the positive direction of the Y axis through the equilibrium position. Try to write this simple and harmonic expression.

Solution: The wave propagates forward along the X axis without attenuation, so the expression of simple harmonic is.

Among them;

By,, know, get.

m

5-8 As shown in the figure, plane waves propagate in the medium, and the wave velocity u = 10 m? S- 1 propagates in the negative direction of X axis, and the vibration equation at point A is known as [SI].

(1) Write the wave function with point A as the coordinate origin;

(2) Write the wave function with point B, which is 5m away from point A, as the coordinate origin;

Solution: (1) m

(2) From the wave function in (1), x=-5 is brought into the above formula, and the initial phase of the particle at point B is obtained as follows.

m

Figure 5-9 shows the waveform of a plane harmonic at time t =0 s, the amplitude of the wave is 0.20 m, and the period is 4.0 s. Find the vibration equation of the particle at the origin of (1) coordinates. (2) If OP=5.0m, write the wave function; (3) Write the vibration equation of point P in the figure.

Solution: It can be seen that the particle at the origin of t=0 is in its equilibrium position and moves in the positive direction of the displacement axis, so.

(1), and the vibration equation of the particle at the coordinate origin is m.

(2) As can be seen from the figure, the wave propagates forward along the X axis, so the wave function is

m

(3) Substitute the coordinate x of point P = 0.5m into the above formula to obtain the vibration equation of point P as follows.

m

5- 10 It is known that the phase difference of the waves emitted by two coherent wave sources is? , the wave path difference to the intersection point p is twice the half wavelength, then the synthesis of point p is [B]

(A) always strengthen

(b) always getting weaker.

(3) Sometimes it is strengthened, sometimes it is weakened, and it changes periodically.

(d) Sometimes it is strengthened, sometimes it is weakened, and there is no certain law.

5- 1 1 As shown in the figure, a simple harmonic propagates in the direction of BP, and the vibration equation caused by it at point B is. The other harmonic propagates in the direction of CP, and the vibration equation caused by it at point C is. Point P is 0.40 m away from point B and 0.50 m away from point C. The wave velocities are all u = 0.20 m? S- 1. Then the phase difference between the two waves at p is 0.

A: Bicycles.

5- 12 As shown in the figure, S 1 and S2 are two coherent wave sources, their vibration directions are perpendicular to the graph, and they emit simple harmonics with wavelengths, and point P is a point in the intersection area of the two waves. It is known that two waves destructively interfere at point P. If the vibration equation of S 1 is 0, the vibration equation of S2 is [].

(1); (b) and:

(c) and: (D) 10.

The answer is (d).

Solution: Let the vibration of S2 be, and the phase difference of two waves at point P be.

, take k=0 or-1, and get.

5- 13 As shown in the figure, S 1 and S2 are two plane harmonic coherent wave sources. Is the phase of S2 ahead of that of S 1? /4, wavelength? = 8.00 m, r 1 = 12.0 m, r2 = 14.0 m, the amplitude caused by S 1 at point P is 0.30 m, and the amplitude caused by S2 at point P is 0.20m m. Find the synthetic amplitude at point P. 。

Solution:

m