Simple Harmonic Motion and Mechanical Waves

Question 1

Question 1: 1. Sound waves belong to mechanical waves, the following description of sound waves is correct

1. Sound waves belong to mechanical waves, the following description of sound waves is correct

A. A. The same column of sound waves has the same wavelength in various media
B. B. The higher the frequency of a sound wave, the faster it travels through the air
C. C. The distance traveled by a vibrating plasma element in the direction of wave propagation in one cycle is equal to one wavelength
D. D. Ultrasonic waves ( $> 20000 \mathrm {~Hz}$) are not prone to significant diffraction when encountering physics exam paper size obstacles in air

Answer

Answer: D

Solution: A. In different media, the wave speed is different, but the frequency is the same, so the wavelength is different, so A error B. Sound waves propagate at the same speed in the same medium, so B error C. Any vibration of the mass is only in their own equilibrium position near the vibration, and does not move forward with the wave, so C error D. Wave diffraction conditions are: the width of the hole, seam or obstacle size and wavelength compared to about the same or smaller than the wavelength, the wavelength of ultrasound is much smaller than the volume, so it is not easy to diffract significantly, so D is correct

Question 2

Question 2: 2. Which of the following statements are true about the relationship between displacement, velocity,...

2. Which of the following statements are true about the relationship between displacement, velocity, and acceleration in simple harmonic motion

A. A. Acceleration increases as displacement decreases and speed increases
B. B. The direction of displacement is always the opposite of the direction of acceleration and the same as the direction of velocity.
C. C. When the object moves toward the equilibrium position, the direction of velocity is opposite to the direction of displacement; when it moves back toward the equilibrium position, the direction of velocity is the same as the direction of displacement.
D. D. The object moves toward the equilibrium position in decelerated motion and away from the equilibrium position in accelerated motion

Answer

Answer: C

Solution: A. When the displacement decreases, the acceleration $a = - \frac { k x } { m }$ also decreases and the velocity increases, so A is wrong; B. Acceleration $a = - \frac { k x } { m }$, the negative sign means that the direction of acceleration is always opposite to the direction of displacement, and it is accelerated during the process of approaching the equilibrium position, and the direction of displacement is opposite to the direction of velocity, so B is wrong; C. When the direction of motion of the object points to the equilibrium position, the direction of velocity is opposite to the direction of displacement; when it turns back to the equilibrium position, the direction of velocity is the same as the direction of displacement, so C is correct; D. When the object moves toward the equilibrium position, it accelerates, and when it moves away from the equilibrium position, it decelerates, so D is wrong. Therefore, choose C.

Question 3

Question 3: 7. analyze the following physical phenomena: (1) "empty mountains do not see people, but hear people...

7. analyze the following physical phenomena: (1) "empty mountains do not see people, but hear people's voices"; (2) color ultrasound to measure the change in the frequency of the reflected wave, so as to know the blood flow rate; (3) around the sound of the sound of the double-stranded tuning fork to walk around, hear the sound of the sound is strong and weak; (4) sonar system, used to detecting objects in the sea, each of these physical phenomena belongs to the wave

A. A. Diffraction, Doppler effect, interference, reflection
B. B. Diffraction, interference, Doppler effect, refraction
C. C. Refraction, interference, Doppler effect, reflection
D. D. Diffraction, refraction, Doppler effect, interference

Answer

Answer: A

Solution: "Empty mountains do not see people, but hear the sound of human speech" belongs to the wave diffraction; color ultrasound to measure the change in the frequency of the reflected wave, so as to know the blood flow rate, belongs to the Doppler effect; around the sound of the double-stranded tuning fork walk around the sound of the sound of the sound of the sound of the sound of the weak, belongs to the interference of the wave; sonar system, used to detect the objects in the sea, belongs to the reflection of the wave.

Question 4

Question 4: 8. With respect to mechanical vibration, the following statements are false

8. With respect to mechanical vibration, the following statements are false

A. A. The direction of the restoring force is always opposite to the direction of displacement
B. B. The direction of acceleration is always opposite to the direction of displacement
C. C. Sometimes the direction of velocity is the same as the direction of displacement, sometimes it is opposite to the direction of displacement
D. D. Mechanical vibration is a uniformly variable linear motion

Answer

Answer: D

Solution: AB. According to the formula for restoring force and Newton's second law: $$ \begin{aligned} & F = - k x \\ & F = m a \end{aligned} $$ It can be seen that the direction of the restoring force and acceleration is always opposite to the direction of displacement, so AB is correct; C. When the object moves from the equilibrium position to the maximum displacement, the direction of the object's velocity is the same as the direction of displacement, and when the object moves from the maximum displacement to the equilibrium position, the direction of the object's velocity is opposite to the direction of displacement, so C is correct; D. The object of mechanical vibration is subjected to variable force, according to Newton's second law can be known that the acceleration also changes, so it is not a uniformly variable linear motion, so D is wrong.

Question 5

Question 5: 9. With respect to mechanical waves, one of the following statements is correct

9. With respect to mechanical waves, one of the following statements is correct

A. A. "Hear the sound before you see it" is a phenomenon that occurs when sound waves interfere.
B. B. The direction of vibration of a point of mass in a transverse wave is co-linear with the direction of propagation of the wave
C. C. The magnitude of the speed of a mechanical wave traveling through a medium is medium dependent
D. D. Determining the direction of travel of a train from the sound of a train whistle utilizes the diffraction of sound waves

Answer

Answer: C

Solution: A. "Before seeing the person, first heard the sound" is a phenomenon produced by diffraction of sound waves, so A error; B. The direction of vibration of the point of physics and the direction of wave propagation perpendicular to the definition of transverse waves, so B error; C. The speed of wave propagation is determined by the medium, independent of the frequency, so C is correct; D. From the sound of the train whistle to determine the direction of travel of the train using the Doppler effect, so D is wrong; Therefore, choose C.

Question 6

Question 6: Figure A is a series of simple harmonic transverse waveform at a certain moment, the point $M , N , ...

Figure A is a series of simple harmonic transverse waveform at a certain moment, the point $M , N , P , Q$ are located in the medium $x = 3 \mathrm {~m} , x = 4 \mathrm {~m} , x = 5 \mathrm {~m} , x = 10 \mathrm {~m}$. The transverse wave propagates to the point $P$ at that moment, and the vibration image of the point $M$ from that moment is shown in Figure B. The following statements are correct ![](/images/questions/phys-waves/image-001.jpg) A ![](/images/questions/phys-waves/image-002.jpg) B

A. A. The speed of propagation of this wave in this medium is $1.25 \mathrm {~m} / \mathrm { s }$
B. B. Wave source oscillation direction is along the $y$ axis.
C. C. The distance traveled by the point $Q$ to the point $P$ is 5 meters.
D. D. When the mass $Q$ vibrates, it vibrates at the same pace as the mass $N$.

Answer

Answer: B

Solution: A. According to the image, the wavelength is 4 m, the period is 4 s $$ v = \frac { \lambda } { T } = 1 \mathrm {~m} / \mathrm { s } $$ B. At this moment the wave has just reached $P$; B. At this moment, the wave just reaches point $P$, and according to the up and down slope method, the vibration direction is along the positive direction of the $y$ axis, and therefore the vibration direction of point $P$ at this moment is the vibration direction of the wave source, so B is correct; C. The wave propagates to point $Q$ at a time that satisfies the following criteria $$ t = \frac { X } { V } = 5 \mathrm {~s} = \frac { 5 } { 4 } T $$ Therefore, the distance from point $P$ is $$ t = \frac { X } { V } = 5 \mathrm {~s} = \frac { 5 } { 4 } T $$. $$ s = 5 \mathrm {~A} = 50 \mathrm {~cm} $$ C is wrong ; D. The distance between the point of mass $Q$ and the point of mass $N$ is $$ \triangle x = 6 m = \frac { 3 } { 2 } \lambda $$ Therefore, the steps are reversed and D is incorrect.

Question 7

Question 7: 11. With respect to the description related to mechanical vibration, the following statements are co...

11. With respect to the description related to mechanical vibration, the following statements are correct

A. A. The period of a single pendulum decreases as the mass of the pendulum ball increases
B. B. The energy of a horizontally placed spring oscillator undergoing simple harmonic vibration is equal to the kinetic energy of the oscillator at the equilibrium position.
C. C. When the pendulum moves to the equilibrium position, the velocity is maximum, the restoring force is zero, and the combined force is zero as well.
D. D. The frequency of the forced vibration is equal to the frequency of the driving force only when resonance occurs

Answer

Answer: B

Solution: A. According to the formula for the period of a single pendulum $$ T = 2 \pi \sqrt { \frac { L } { g } } $$ A. According to the formula $$ T = 2 \pi \sqrt { \frac { L } { g } } $$, the period of the pendulum has nothing to do with the mass of the pendulum ball; B. The equilibrium position of the horizontally placed spring oscillator doing simple harmonic vibration is located in the original length of the spring, the position of the elastic potential energy is zero, the maximum kinetic energy, the spring oscillator doing simple harmonic vibration when the mechanical energy is conserved, so the horizontal placement of the spring oscillator doing simple harmonic vibration of the energy is equal to the equilibrium position of the oscillator when the kinetic energy, B is correct; C. The pendulum moves to the equilibrium position, the speed is maximum, the return force is zero, due to the pendulum to do circular motion, radial force to provide centripetal force, so the combined force is not zero, C is wrong; D. According to the characteristics of forced vibration, the frequency of forced vibration must be equal to the frequency of the external driving force, D error.

Question 8

Question 8: 12. A spring oscillator placed on a smooth horizontal surface starts to vibrate after the spring is ...

12. A spring oscillator placed on a smooth horizontal surface starts to vibrate after the spring is compressed $x$ for the first time and after the spring is compressed $2 x$ for the second time, then the ratio of the amplitudes of the two vibrations is

A. A. $1 : 2$
B. B. 2: 1
C. C. $1 : 4$
D. D. $4 : 1$

Answer

Answer: A

Solution: Because the maximum distance from the equilibrium position during the vibration of the spring oscillator is the amplitude, then by the meaning of the question, the compression of the spring in this scenario is the amplitude, so the ratio of the amplitude of the two vibrations is $1 : 2$.

Question 9

Question 9: 13. The waveform of a simple harmonic transverse wave propagating in the positive direction along th...

13. The waveform of a simple harmonic transverse wave propagating in the positive direction along the $x$ axis at $t _ { 1 } = 0$ is shown in the figure, and at this time, the wave propagates to the point $x = 2.0 \mathrm {~m}$ at the point $B$, and the point $A$ is located exactly at the trough of the wave. FORMULA_4]] is located exactly at the trough of the wave. When $t _ { 2 } = 0.6 \mathrm {~s}$, the prime $A$ happens to be in the crest position for the second time, the following statement is correct ![](/images/questions/phys-waves/image-003.jpg)

A. A. At the $t _ { 1 } = 0$ moment, the point $B$ is moving in the positive direction of the y-axis.
B. B. After another quarter of a cycle, the mass $A$ will reach the mass $B$
C. C. The period of this wave train is 0.4 s
D. D. The wave speed of this train is $10 \mathrm {~m} / \mathrm { s }$

Answer

Answer: C

Solution: A. By the waveform diagram, $t _ { 1 } = 0$ moment, the point $B$ along the y-axis negative direction of motion, option A error; B. In the process of mechanical wave propagation, the particle can only vibrate near its equilibrium position and does not migrate with the wave, option B is wrong; C. When $t _ { 2 } = 0.6 \mathrm {~s}$, the point $A$ is exactly in the peak position for the second time. $$ 1.5 T = 0.6 \mathrm {~s} $$ Then the period of this wave train is $$ T = 0.4 \mathrm {~s} $$ Option C is correct; D. The speed of the wave is $$ v = \frac { \lambda } { T } = \frac { 2 } { 0.4 } \mathrm {~m} / \mathrm { s } = 5 \mathrm {~m} / \mathrm { s } $$ Option D is wrong.

Question 10

Question 10: A simple harmonic transverse wave propagates along the $x$ axis in the positive direction, and the w...

A simple harmonic transverse wave propagates along the $x$ axis in the positive direction, and the waveform at the $t = 0$ time is shown by the solid line in the figure, and the waveform at the $t = 0.1 \mathrm {~s}$ time is shown by the dashed line in the figure. The source of the wave is not at the origin $O , P$ but is a mass point in the propagation medium at a distance from the origin $\mathrm { xP } = 2.5 \mathrm {~m}$. The following statement is correct ![](/images/questions/phys-waves/image-004.jpg)

A. A. The amplitude of the mass point $P$ is 0.05 m
B. B. The frequency of the wave may be 7.5 Hz.
C. C. The wave propagation speed may be $50 \mathrm {~m} / \mathrm { s }$
D. D. A point 5 m away from $P$ at the moment $t = 0.1 \mathrm {~s}$ must be moving in the positive direction of the $y$ axis.

Answer

Answer: C

Solution: A. Amplitude is the maximum distance of the point from the equilibrium position, from the figure, the amplitude of the point $P$ is 0.1 m, so A is wrong. B. The wave propagates in the positive direction along the $x$ axis. $$ t = n T + \frac { T } { 4 } \quad ( n = 0,1,2,3 , \cdots ) $$ The period is $$ T = \frac { 4 t } { 4 n + 1 } = \frac { 0.4 } { 4 n + 1 } \mathrm {~s} \quad ( n = 0,1,2,3 , \cdots ) $$ The frequency is $$ f = \frac { 1 } { T } = \frac { 4 n + 1 } { 0.4 } \mathrm {~Hz} = ( 10 n + 2.5 ) \mathrm { Hz } ( n = 0,1,2,3 , \cdots ) $$ So the frequency of the wave may be $2.5 \mathrm {~Hz} , 12.5 \mathrm {~Hz} , \cdots$, not 7.5 Hz, so B is wrong; C. The wave speed is $$ v = \lambda f = 4 \times ( 10 n + 2.5 ) \mathrm { m } / \mathrm { s } = ( 40 n + 10 ) \mathrm { m } / \mathrm { s } ( n = 0,1,2,3 , \cdots ) $$ C. The wave speed is $n = 1$ when $v = 50 \mathrm {~m} / \mathrm { s }$ is $n = 1$, so C is correct. D. The wave propagates along the $x$ axis in the positive direction, and at the $t = 0.1 \mathrm {~s}$ time $P$ is moving in the positive direction along the $y$ axis, and the mass points 5 m apart are at 7.5 m and - 2.5 m respectively. The mass at 7.5 m vibrates in the same positive direction along the $y$ axis as the mass at 3.5 m, while the mass at -2.5 m from the origin vibrates in the same negative direction along the $y$ axis as the mass at 1.5 m. D is wrong.

Question 11

Question 11: 15. Analyze the following physical phenomena: (1) water waves in a pond continue to propagate around...

15. Analyze the following physical phenomena: (1) water waves in a pond continue to propagate around obstacles; (2) thunder booms endlessly in the clouds; (3) color ultrasound measures the change in the frequency of the reflected wave, so as to know the flow rate of the blood; (4) headphones with an active noise-canceling function, according to the reception of the (4) headphones with active noise-canceling function, according to the collected noise information to send out a specific sound wave can cancel out the noise. These physical phenomena belong to the wave

A. A. Doppler effect, interference phenomena, diffraction phenomena, reflection phenomena
B. B. Diffraction phenomena, reflection phenomena, Doppler effect, interference phenomena
C. C. Doppler effect, reflection phenomena, diffraction phenomena, interference phenomena
D. D. Diffraction phenomena, interference phenomena, Doppler effect, reflection phenomena

Answer

Answer: B

Solution: (1) water waves in a pond continue to propagate around obstacles belong to the phenomenon of diffraction; (2) thunder in the clouds roar unceasingly belong to the phenomenon of reflection; (3) color ultrasound to measure the change in the frequency of the reflected wave, so as to know the flow rate of the blood belongs to the Doppler effect; (4) with the active noise-canceling function of the earphones, according to the collected noise information to send out a specific sound wave can cancel the noise belongs to the phenomenon of interference. Therefore, choose B.

Question 12

Question 12: 16. The figure shows the waveforms generated by two wave sources $S _ { 1 } , S _ { 2 }$ in a sink, ...

16. The figure shows the waveforms generated by two wave sources $S _ { 1 } , S _ { 2 }$ in a sink, where the solid line shows the peaks and the dashed line shows the troughs. In order for the two waves to interfere in the region where they meet, they should be ![](/images/questions/phys-waves/image-005.jpg)

A. A. Increase the distance between $S _ { 1 }$ and $S _ { 2 }$.
B. B. Decrease the vibration period of $S _ { 1 }$.
C. C. Replace the water in the tank with oil
D. D. Increase the vibration frequency of $S _ { 2 }$.

Answer

Answer: B

Solution: BD. The medium is the same, then the wave propagation speed is certain, according to the figure can be seen, the wave source S1 propagation wavelength is larger, according to $v = \frac { \lambda } { T } = \lambda f$ According to $v = \frac { \lambda } { T } = \lambda f$, the frequency of the wave source $S _ { 1 }$ is smaller and the period is larger. In order to have a stable interference pattern, it is necessary to make the frequency or period of the two sources equal, that is, to reduce the period of the vibration of $S _ { 1 }$ or to reduce the frequency of the vibration of $S _ { 2 }$, so B is correct. D is wrong; AC. In order to produce a stable interference pattern, the frequency or period of the two wave sources need to be equal, independent of the spacing between $S _ { 1 }$ and $S _ { 2 }$ and the medium in the slot, so AC is wrong.

Question 13

Question 13: 17. The wavelength of a wave in air is 40 cm and the velocity is $340 \mathrm {~m} / \mathrm { s }$,...

17. The wavelength of a wave in air is 40 cm and the velocity is $340 \mathrm {~m} / \mathrm { s }$, when refracted into medium 1, the wavelength changes to 100 cm, and the following statements are true

A. A. 8.5 Hz in air
B. B. 85 Hz in air
C. C. The propagation speed in medium 1 is $85 \mathrm {~m} / \mathrm { s }$
D. D. The propagation speed in medium 1 is $850 \mathrm {~m} / \mathrm { s }$

Answer

Answer: D

Solution: AB . The frequency in air is $$ f = \frac { v } { \lambda } = \frac { 340 } { 0.4 } \mathrm {~Hz} = 850 \mathrm {~Hz} $$ Option AB is incorrect ; CD. the frequency remains the same when it enters medium 1, then the speed of propagation is $$ v ^ { \prime } = \lambda f = 1 \times 850 \mathrm {~m} / \mathrm { s } = 850 \mathrm {~m} / \mathrm { s } $$ Option C is wrong, D is correct.

Question 14

Question 14: 19. Coordinates at the origin of the wave source to do simple harmonic motion, it is formed in a uni...

19. Coordinates at the origin of the wave source to do simple harmonic motion, it is formed in a uniform medium of simple harmonic transverse wave propagation along the $x$ axis in the positive direction, the source of the wave vibration 4 s after the wave is just transmitted to the $x = 12 \mathrm {~m}$ place, the waveform is shown in the figure. The following statement is correct ( ) ![](/images/questions/phys-waves/image-006.jpg)

A. A. The $M$ point is vibrating in the negative direction along the $y$ axis.
B. B. The magnitude of the wave's velocity in the medium is $3 \mathrm {~m} / \mathrm { s }$
C. C. The equilibrium position of the mass point $M$ is at $x = 4 \mathrm {~m}$.
D. D. The wave has a propagation period of 4 s

Answer

Answer: B

Solution: A. The wave propagates in the positive direction along the $x$ axis, and according to the "same side method", the $M$ point in the figure is vibrating in the positive direction along the $y$ axis, so A is wrong; B. According to $v = \frac { \Delta x } { \Delta t } = \frac { 12 } { 4 } \mathrm {~m} / \mathrm { s } = 3 \mathrm {~m} / \mathrm { s }$, so B is correct; C. The points on the transverse wave vibrate back and forth near their equilibrium positions, perpendicular to the direction of wave propagation, and the equilibrium position of point M is to the left of $x = 4 \mathrm {~m}$, so C is wrong; D. From the figure, the wave has a wavelength of 8 m and a period of $T = \frac { \lambda } { v } = \frac { 8 } { 3 } \mathrm {~s}$. D. The wave has a period of $T = \frac { \lambda } { v } = \frac { 8 } { 3 } \mathrm {~s}$.

Question 15

Question 15: 20. An oscillator moves in simple harmonic motion along the $x$ axis with its equilibrium at the ori...

20. An oscillator moves in simple harmonic motion along the $x$ axis with its equilibrium at the origin. The displacement of the oscillator at $t = 0$ is $- 0.2 \mathrm {~m} , ~ t = 1 \mathrm {~s}$. The displacement of the oscillator at $- 0.2 \mathrm {~m} , ~ t = 1 \mathrm {~s}$ is 0.2 m. Then ( )

A. A. If the amplitude is ${ } _ { 0.2 \mathrm {~m} }$, the period of the oscillator may be $\frac { 1 } { 3 } \mathrm {~s}$
B. B. If the amplitude is ${ } _ { 0.2 \mathrm {~m} }$, the period of the oscillator may be $\frac { 4 } { 5 } \mathrm {~s}$
C. C. If the amplitude is 0.4 m, the period of the oscillator may be $\frac { 6 } { 11 } \mathrm {~s}$
D. D. If the amplitude is 0.4 m, the period of the oscillator may be $\frac { 5 } { 13 } \mathrm {~s}$

Answer

Answer: C

Solution: AB. If the amplitude is 0.2 m, then $$ n T + \frac { T } { 2 } = 1 \mathrm {~s} \quad ( n = 0,1,2,3 \cdots ) $$ then $$ T = \frac { 2 } { 2 n + 1 } \mathrm {~s} ( n = 0,1,2,3 \cdots ) $$ The period of the oscillator cannot be $\frac { 1 } { 3 } \mathrm {~s}$ or $\frac { 4 } { 5 } \mathrm {~s}$, option AB is wrong; CD. If the amplitude is 0.4 m, according to $$ 0.4 \times \sin \left( \frac { 2 \pi } { T } \times \frac { T } { 12 } \right) \mathrm { m } = 0.2 \mathrm {~m} $$ then we have $$ n T + \frac { T } { 2 } = 1 \mathrm {~s} \quad ( n = 0,1,2,3 \ldots ) $$ can be obtained $$ T = \frac { 2 } { 2 n + 1 } \mathrm {~s} ( n = 0,1,2,3 \ldots ) $$ Or $$ n T + \frac { T } { 6 } = 1 \mathrm {~s} ( n = 0,1,2,3 \ldots ) $$ can be obtained $$ T = \frac { 6 } { 6 n + 1 } \mathrm {~s} ( n = 0,1,2,3 \ldots ) $$ Or $$ n T + \frac { 5 T } { 6 } = 1 \mathrm {~s} \quad ( n = 0,1,2,3 \ldots ) $$ can be obtained $$ T = \frac { 6 } { 6 n + 5 } \mathrm {~s} ( n = 0,1,2,3 \ldots ) $$ which is found in the $$ T = \frac { 6 } { 6 n + 5 } \mathrm {~s} ( n = 0,1,2,3 \ldots ) $$ expression, when $n = 1$ $$ T = \frac { 6 } { 11 } \mathrm {~s} $$ But it is impossible to equal $\frac { 5 } { 13 } \mathrm {~s}$ in all three expressions, option C is correct and D is wrong.

Question 16

Question 16: 21. A spring oscillator is in simple harmonic motion in the horizontal direction with period $T$, th...

21. A spring oscillator is in simple harmonic motion in the horizontal direction with period $T$, then the following statement is correct ( )

A. A. If the magnitude of the displacement of the oscillator in ${ } _ { \Delta t }$ time is equal to one amplitude, then ${ } _ { \Delta t }$ must be equal to an integer multiple of $\frac { T } { 4 }$.
B. B. If the speed of the oscillator is the same at ${ } _ { t }$ and ${ } _ { t + \Delta t }$, then ${ } _ { \Delta t }$ must be equal to an integer multiple of $\frac { T } { 2 }$.
C. C. If $\Delta t = \frac { T } { 2 }$, then the lengths of the springs must be equal at ${ } _ { t }$ and $t + \Delta t$.
D. D. If $\Delta t = \frac { T } { 2 }$, then the magnitude of the acceleration of the oscillator must be equal at the ${ } _ { t }$ and ${ } _ { t + \Delta t }$ moments.

Answer

Answer: D

Solution: A. Since the speed of the oscillator near the equilibrium position is greater than the speed away from the equilibrium position, then in the space near the equilibrium position of the left and right sides of the movement process, if ${ } _ { \Delta t }$ time the displacement of the oscillator is equal to the size of an amplitude, then ${ } _ { \Delta t }$ is smaller than $\frac { T } { 4 }$, namely [INLINE_FORMULA_3]] must be equal to an integer multiple of $\frac { T } { 4 }$, so A is wrong; B. According to the symmetry of simple harmonic motion, when $\Delta t$ is equal to an odd multiple of $\frac { T } { 2 }$, the oscillator arrives at the symmetry of the relative equilibrium position, and the velocity size of the oscillator is equal to that of the time of $t$, and they are in opposite directions, so B is wrong. B is wrong; C. If $\Delta t = \frac { T } { 2 }$, then the oscillator just reached the symmetry point relative to the equilibrium position, this time relative to the equilibrium position of the size of the displacement C. If $\Delta t = \frac { T } { 2 }$, the oscillator just reaches the symmetry point relative to the equilibrium position, then the displacements relative to the equilibrium position are equal in magnitude and opposite in direction, and the lengths of the springs are not equal at the $t$ and $t + \Delta t$ moments; D. If $\Delta t = \frac { T } { 2 }$, the oscillator just reaches the symmetry point relative to the equilibrium position, then the acceleration of the oscillator at the ${ } _ { t }$ and $t + \Delta t$ moments must be equal, so D is correct.

Question 17

Question 17: 22. As shown in the figure represents two columns of the same frequency coherent water waves in the ...

22. As shown in the figure represents two columns of the same frequency coherent water waves in the moment of superposition, the solid line in the figure indicates the wave peaks, dashed line indicates the wave valley, known as two columns of the amplitude of the wave are 1 cm, the wave speed is $2 \mathrm {~m} / \mathrm { s }$, the wavelength of 0.2 m, the following statement is correct () ![](/images/questions/phys-waves/image-007.jpg)

A. A. Points B and D are points where the vibration is intensified and the displacement is always at its maximum.
B. B. Points A and C are points where the vibration is weakened.
C. C. The difference in height between points B and D at that moment is 2 cm.
D. D. $\mathrm { t } = 0.025 \mathrm {~s}$ Time point A is 2 cm from the equilibrium position.

Answer

Answer: B

Solution: Point B is the peak and crest, and point D is the superposition of the trough and trough, both of which are vibration intensification points, with maximum amplitude, but the displacement is sometimes large and sometimes small. Therefore, A is wrong. Point A and point C are superposition of wave crest and trough, which is a point of weakening of vibration, so B is correct; point D is superposition of wave crest and wave peak at that moment, and vibration is strengthened, and the displacement away from the equilibrium position is $2 \mathrm {~cm} , \mathrm {~B}$ The point is superposition of wave trough and wave valley at that moment, and vibration is strengthened, and the displacement away from the equilibrium position is - 2 cm, then $\mathrm { D } , \mathrm {~B}$ The difference in height between the two points at that moment is 4 cm; so C is wrong. The period of wave propagation is $T = \frac { \lambda } { v } = \frac { 0.2 } { 2 } \mathrm {~s} = 0.1 \mathrm {~s}$, and then $\mathrm { t } = 0.025 \mathrm {~s}$ is $1 / 4$ period, point A is the meeting of the wave peak and the wave valley, so the displacement away from the equilibrium is 0. Therefore, D is wrong. Therefore, D is wrong. Therefore, B is wrong.

Question 18

Question 18: 23. When a train of waves passes from one medium into another ()

23. When a train of waves passes from one medium into another ()

A. A. Wave speed frequency and wavelength are constant
B. B. Wave speed is constant, frequency and wavelength change
C. C. Frequency remains constant, wave speed and wavelength change
D. D. Wave speed, frequency and wavelength all change

Answer

Answer: C

Solution: The frequency of a wave is determined by the wave source, when the wave goes from one medium to another, the frequency remains unchanged. Wave speed is determined by the medium, different media, wave speed is different, by the formula $v = \lambda f$, $f$ unchanged, $v$ change, then $\lambda$ change, so C is correct. Therefore, C is correct.

Question 19

Question 19: 24. A particle moves in simple harmonic motion, and its displacement $x$ is related to the time t by...

24. A particle moves in simple harmonic motion, and its displacement $x$ is related to the time t by the curve shown in the figure. From the figure, we can see that at $t = 3 \mathrm {~s}$, the particle ( ) [IMAGE_2]]

A. A. Velocity is a positive maximum, acceleration is zero
B. B. Negative maximum velocity, zero acceleration
C. C. Zero velocity, positive maximum acceleration
D. D. Zero velocity and negative acceleration maxima

Answer

Answer: A

Solution: At $t = 3 \mathrm {~s}$, the displacement of the mass is 0, it is in equilibrium, the acceleration is zero, and the velocity of the mass is positively maximum

Question 20

Question 20: 25. The following physical phenomena are correctly stated ( )

25. The following physical phenomena are correctly stated ( )

A. A. The phenomenon of "hearing but not seeing" is caused by the interference of sound waves.
B. B. To prevent the occurrence of object resonance, try to make the frequency of the driving force coincide with the intrinsic frequency of the object
C. C. When a car crosses a row of equally spaced speed bumps, the greater the speed, the greater the amplitude
D. D. The medical use of "color ultrasound" to measure blood flow velocity utilizes the Doppler effect of ultrasound.

Answer

Answer: D

Solution: A. The phenomenon of "hearing the sound but not seeing the person" is caused by the diffraction of sound waves, option A is wrong; B.When utilizing resonance, try to make the frequency of the driving force coincide with the intrinsic frequency of the object, option B is wrong; C. When a car passes through a row of equally spaced speed bumps, there are $$ T = \frac { x } { v } = \frac { 1 } { f } $$ The greater the speed of the car the greater the frequency of the driving force, but do not know the intrinsic frequency of the car, so it is impossible to determine whether the amplitude is greater, option C error ; D. The medical use of "color ultrasound" way to measure blood flow velocity, is the use of ultrasound Doppler effect, option D is correct.

Question 21

Question 21: 26. As shown in the figure, two students respectively pull the ends of a 1.2 m long rope $\mathrm { ...

26. As shown in the figure, two students respectively pull the ends of a 1.2 m long rope $\mathrm { A } , \mathrm {~B} , ~ t = 0$ moment, the two students at the same time shake the ends of the rope, so that A, B began to do simple harmonic vibration in the direction of the firm straight, resulting in the propagation of two columns of waves along the rope, the source of vibration for the wave speed of A is ${ } ^ { ~ } { } ^ { ~ 1 }$, vibration for the source of the wave speed of B is ${ } ^ { ~ } { } ^ { ~ 2 }$. [INLINE_FORMULA_2]]. $t = 0$ At 4 s, the two columns of waves propagate to exactly two points of $P , Q$, and the waveforms are shown in the figure, then ( ) ![](/images/questions/phys-waves/image-009.jpg)

A. A. The two wave trains oscillate in opposite directions
B. B. $v _ { 1 } = 2 v _ { 2 }$
C. C. The two wave trains have the same period
D. D. After 0.8 s, the two waves meet for the first time.

Answer

Answer: D

Solution: A. At the time shown in the figure, the two columns of waves propagate to the two points of $P , Q$, and according to the method of upward and downward slopes, we can get that the vibration direction of the two points of $P , Q$ is firm and straight upward, so the two columns of waves have the same direction of vibration, so A is wrong; B. The propagation speed of mechanical waves is related to the medium, because in the same medium propagation, so the propagation speed is the same, so B error; C. The vibration period of the point $A$ is $$ T _ { \mathrm { A } } = 2 t = 0.8 \mathrm {~s} $$ The period of vibration of point B is $$ T _ { \mathrm { A } } = 2 t = 0.8 \mathrm {~s} $$. $$ T _ { \mathrm { B } } = t = 0.4 \mathrm {~s} $$ Therefore, C is wrong; D.The two trains of waves travel at the same speed both $$ v = \frac { x } { t } = \frac { 0.2 } { 0.4 } \mathrm {~m} / \mathrm { s } = 0.5 \mathrm {~m} / \mathrm { s } $$ From the figure, we can see that the two waves meet for the first time at $x = 0.6 \mathrm {~m}$, then the time from the moment shown in the figure to the first encounter is $$ \Delta t = \frac { \Delta x } { v } = \frac { 0.6 - 0.2 } { 0.5 } \mathrm {~s} = 0.8 \mathrm {~s} $$ Therefore, D is correct.

Question 22

Question 22: 27. As shown in Figure A is a series of simple harmonic transverse waves propagating along the $x$ a...

27. As shown in Figure A is a series of simple harmonic transverse waves propagating along the $x$ axis in the $t = 2 \mathrm {~s}$ time of the wave image, $a , b , c$ for the medium of the three mass points, Figure B indicates that the wave $x = 6 \mathrm {~m}$ at the $a$ point of vibration image, the following statement is correct ( ). INLINE_FORMULA_4]] at $x = 6 \mathrm {~m}$, the following statements are correct ( ) ![](/images/questions/phys-waves/image-010.jpg) A ![](/images/questions/phys-waves/image-011.jpg) B

A. A. The wave propagates negatively along the $x$ axis.
B. B. At the $t = 4 \mathrm {~s}$ moment, the velocity of the vibration of the point $b$ is in the same direction as the acceleration.
C. C. At the $t = 3.5 \mathrm {~s}$ moment, the $c$ vibrates in the direction of the velocity along the $y$ axis.
D. D. Relationship between the displacement of the point $c$ and time $y = 10 \sin \left( \frac { \pi t } { 2 } - \frac { 5 } { 6 } \pi \right) \mathrm { cm }$

Answer

Answer: BD

Question 23

Question 23: 28. The following statements are true about the phenomenon of waves

28. The following statements are true about the phenomenon of waves

A. A. Traffic police use speed cameras to transmit ultrasonic waves of known frequency to moving vehicles, according to the frequency change of the reflected wave to determine the speed of the vehicle, which is the use of the Doppler effect of waves
B. B. "Hearing is not seeing" is a reflection of sound waves.
C. C. Doctors transmit ultrasound waves of known frequency into the body, and determine the speed of blood flow based on the change in frequency of the received ultrasound waves that are reflected by the blood flow in the blood vessels by using the phenomenon of wave interference
D. D. The incessant roar of summer thunder is a diffraction of sound waves

Answer

Answer: A

Solution: A. Traffic police use speed cameras to transmit ultrasonic waves of known frequency to moving vehicles, according to the frequency change of the reflected wave to determine the speed of the vehicle, which is the use of the Doppler effect of the wave, option A is correct; B. "hear their voices do not see their people" is the diffraction phenomenon of sound waves, option B is wrong; C. Doctors to the human body to launch a known frequency of ultrasound, according to the received by the blood flow in the blood vessels after the ultrasound frequency changes, to determine the speed of blood flow is the use of the Doppler phenomenon, option C is wrong; D. Summer thunder roars endlessly, which is the reflection of sound waves, option D error.

Question 24

Question 24: 29. To the prime $Q$ as the source of the mechanical wave on the rope to the prime $P$ when the wave...

29. To the prime $Q$ as the source of the mechanical wave on the rope to the prime $P$ when the waveform is shown in the figure, the following statements are correct ) ![](/images/questions/phys-waves/image-012.jpg)

A. A. The rope wave is a longitudinal wave
B. B. $Q$ The starting direction of the point is facing downward.
C. C. When the $Q$ point stops vibrating, the rope wave does not disappear immediately.
D. D. The point $Q$ may move to the position of the point $P$.

Answer

Answer: C

Solution: A. The direction of vibration of the point in the rope wave is perpendicular to the direction of propagation of the wave, which is a transverse wave, not a longitudinal wave, so A is wrong; B. When the wave reaches the point $P$, the same side method shows that the vibration direction of the point $P$ is upward, because the vibration direction of the wave source is the same as the vibration direction of the point, so the vibration direction of the point $Q$ is upward, so B is wrong; C. After the source stops vibrating, the wave that has been propagated will not disappear immediately, but will continue to propagate, so C is correct; D. In the process of mechanical wave propagation, the point will only vibrate near its equilibrium position and will not migrate with the wave, so it is impossible for the point $Q$ to move to the position of the point $P$, so D is wrong.

Question 25

Question 25: 30. Active noise-canceling headphones are increasingly popular among young people because of their u...

30. Active noise-canceling headphones are increasingly popular among young people because of their unique sense of hearing, and its noise reduction principle as shown in the figure, there are microphones in the headset dedicated to the collection of ambient noise, the headset collects the ambient noise will be produced and the phase of the ambient noise opposite to the noise-canceling sound waves, so as to achieve the effect of noise reduction. The solid line in the figure corresponds to the ambient noise, the dotted line corresponds to the noise reduction sound wave generated by the headset, then the following statement is correct ![](/images/questions/phys-waves/image-013.jpg)

A. A. The noise reduction process applies the Doppler effect of sound waves
B. B. Noise-canceling sound waves and ambient noise sound waves may not travel through the air at the same magnitude.
C. C. Noise reduction sound waves must be equal in frequency to ambient noise sound waves.
D. D. The $P$ point is the point where the vibration is intensified.

Answer

Answer: C

Solution: ABC. As can be seen from the figure, noise reduction sound waves and environmental sound waves of equal wavelength, equal propagation speed in the same medium, then the same frequency, superimposed on the interference phenomenon occurs, due to the two columns of sound waves of the same amplitude, phase reversal, so the vibration is weakened, and play a role in noise reduction, it is AB wrong, C is correct; D. As can be seen from the figure, the two sound waves make the $P$ point vibrate in the opposite direction, so the $P$ point is the point of vibration weakening, so D is wrong.

Question 26

Question 26: 31. The following statements are true about the restoring force in simple harmonic motion

31. The following statements are true about the restoring force in simple harmonic motion

A. A. The restoring force of a simple harmonic motion may be a constant force
B. B. The direction of acceleration of an object in simple harmonic motion may be the same as the direction of displacement
C. C. In $F = - k x$, $k$ is the coefficient of strength of the spring, and $x$ is the length of the spring.
D. D. An object in simple harmonic motion must have zero restoring force each time it passes the equilibrium position.

Answer

Answer: D

Solution: A.According to the definition of simple harmonic motion, when the object is doing simple harmonic motion, the return force is $F = - k x , k$ is the scale factor, not necessarily the strength coefficient of the spring, $x$ is the displacement of the object with respect to the equilibrium position, not the length of the spring, because $x$ is the change of the return force, so A, C is wrong. Since $x$ is variable, the restoring force cannot be constant; B. The direction of the restoring force is always opposite to the direction of displacement. According to Newton's second law, the direction of acceleration must be opposite to the direction of displacement, so B is wrong; D. The return force of an object in simple harmonic motion must be zero every time it passes the equilibrium position, so D is correct.

Question 27

Question 27: 32. The following phenomenon is not an obvious diffraction of a wave

32. The following phenomenon is not an obvious diffraction of a wave

A. A. "The walls have ears.
B. B. "I don't see anyone on the empty mountains, but I hear the sound of people's voices.
C. C. No wave-free area left behind the reeds in the river
D. D. Water waves in front of the rockery in the pool, no water waves behind the rockery

Answer

Answer: D

Solution: A. "Walls have ears" is a wave around the obstacle to continue to propagate, is the diffraction of sound, so A does not meet the meaning of the question; B. "Empty mountains do not see people, but hear the sound of people talking", this is the obstacle blocking the light, can not block the sound, the sound of diffraction obvious, so B. ## B. "I can't see the people on the mountain, but I can hear their voices" is an obstacle blocking light and sound; C. No wave-free area behind the reeds in the river, is the obvious diffraction phenomenon of water waves, so C does not meet the meaning of the question; D. The absence of water waves behind the rockery in the pool indicates that the diffraction phenomenon is not obvious, so D is consistent with the meaning of the question.

Question 28

Question 28: 33 . One of the following statements is correct

33 . One of the following statements is correct

A. A. When utilizing resonance, the frequency of the driving force should be kept at a certain gap with the intrinsic frequency of the vibration system; when preventing resonance, the frequency of the driving force should be close to or equal to the intrinsic frequency of the vibration system
B. B. The students stood side-by-side, with their hands on each other, and rose and squatted in turn, as a way of simulating the propagation of a mechanical longitudinal wave.
C. C. We hear the sirens at a lower and lower pitch as the fire trucks approach and a higher and higher pitch as the fire trucks depart
D. D. Two speakers in a school playground play the same music, and when a person walks on the playground, he feels that some parts are louder and some parts are quieter, which is a wave interference phenomenon

Answer

Answer: D

Solution: A. In the use of resonance, the frequency of the driving force should be close to or equal to the intrinsic frequency of the vibration system; in the prevention of resonance, the frequency of the driving force and the intrinsic frequency of the vibration system should be kept at a certain gap, so A error; B. The students stand side by side, and hand in hand, in order to rise and squat, as a way to simulate the propagation of mechanical transverse waves, so B error ; C. siren fire trucks coming, we hear the siren pitch is getting higher and higher, the fire trucks leave, we hear the siren pitch is getting lower and lower, so C error; D. school playground on the two speakers playing the same music, people walking on the playground, feel the sound of some places, some places sound small, this is the wave interference phenomenon, so D correct.

Question 29

Question 29: Figure A shows a series of simple harmonic transverse wave in $t = 0.2 s$, Figure B is the wave on t...

Figure A shows a series of simple harmonic transverse wave in $t = 0.2 s$, Figure B is the wave on the $A$ vibration image of the mass point, then the following judgment is correct () [IMAGE_2]] A ![](/images/questions/phys-waves/image-015.jpg) B how (what extent)

A. A. The wave speed of this train is $2.5 \mathrm {~m} / \mathrm { s }$
B. B. This wave propagates positively along the $x$ axis.
C. C. If this wave encounters another simple harmonic transverse wave and undergoes stable interference, the frequency of the wave encountered is 2.5 Hz
D. D. If the wave is diffracted by an obstacle, the obstacle must be smaller than 40 cm.

Answer

Answer: C

Solution: A. From the image: $T = 0.4 \mathrm {~s} , \lambda = 0.4 \mathrm {~m}$, the wave speed is $v = \frac { \lambda } { T } = \frac { 0.4 } { 0.4 } \mathrm {~m} / \mathrm { s } = 1 \mathrm {~m} / \mathrm { s }$, so A is wrong . B. From Figure B: $t = 0.2 \mathrm {~s}$ moment point $A$ is vibrating upward through the equilibrium position, according to the translation method of the waveform can be known that the wave propagates along the negative direction of the $x$ axis. Therefore, B is wrong. C. Stable interference requires the same frequency, so the frequency of the wave encountered is $f = 1 / T = 2.5 \mathrm {~Hz}$ to produce stable interference, so C is correct. D. If the wave encounters an obstacle that diffracts significantly, the obstacle must be about the same size or smaller than 40 cm, so D is incorrect.

Question 30

Question 30: The following is the waveform image of a mechanical wave at $t = 0$ and the vibration image of a poi...

The following is the waveform image of a mechanical wave at $t = 0$ and the vibration image of a point $P$ corresponding to the wave, according to the image is correct. ![](/images/questions/phys-waves/image-016.jpg) ![](/images/questions/phys-waves/image-017.jpg)

A. A. $P$ point amplitude is 0
B. B. $Q$ Point velocity toward $y$ axis positive direction
C. C. The wave propagates to the right, $v = 2.0 \mathrm {~m} / \mathrm { s }$
D. D. The wave propagates to the left, $V = 4.0 \mathrm {~m} / \mathrm { s }$

Answer

Answer: C

Solution: A. From the figure, the amplitude $$ A = 10 \mathrm {~cm} $$ amplitude of each mass on the wave is 10 cm , A is wrong; B. At $t = 0$, the point $P$ vibrates upward, so the wave propagates to the right, and then the point $Q$ vibrates downward, i.e., the velocity of the point $Q$ is in the negative direction of the $y$ axis. FORMULA_8]] axis in the negative direction, B is wrong; CD. From the wave image and the vibration image of the point $P$, we can see that BLOCK_FORMULA_1]] is a wave that vibrates in the negative direction of the $P$ axis. According to $$ v = \frac { \lambda } { T } $$ Solve for $$ v = 2 \mathrm {~m} / \mathrm { s } $$ C is correct and D is incorrect.

Question 31

Question 31: 36. As shown in the figure is a rope wave formation process. The simple harmonic motion of point 1 i...

36. As shown in the figure is a rope wave formation process. The simple harmonic motion of point 1 in the straight direction under the action of an external force drives point $2 , 3 , 4 \ldots \ldots$ to vibrate up and down in turn, transmitting the vibration from the left end to the right end of the rope. At $t = \frac { T } { 4 }$, point 5 is just about to start moving. The following statement is correct. ![](/images/questions/phys-waves/image-018.jpg) ![](/images/questions/phys-waves/image-019.jpg)

A. A. At $t = \frac { T } { 4 }$, mass 5 starts to move downward.
B. B. $t = \frac { T } { 4 }$, the direction of acceleration of mass 3 is upward.
C. C. For a short period of time, starting from $t = \frac { T } { 2 }$, the velocity of point 8 is increasing.
D. D. The acceleration of point 8 is increasing for a short period of time starting from $t = \frac { T } { 2 }$.

Answer

Answer: D

Solution: A. Point 1 is the source of the wave, and the wave propagates to the right, $t = \frac { T } { 4 }$ when point 5 starts to vibrate, and the direction of vibration is upward, so A is wrong; B. From the figure, we can see that the displacement of point 3 is positive at $t = \frac { T } { 4 }$, so the direction of the restoring force is downward, and the direction of acceleration is downward, so B is wrong; C. In $t = \frac { T } { 2 }$, the vibration of point 8 is the same as that of point 4 in $t = \frac { T } { 4 }$, so point 8 vibrates upward and its speed decreases in $t = \frac { T } { 2 }$, so C is wrong; D. From ${ } ^ { t = \frac { T } { 2 } \text { 开始的一小段时间内,质点 } 8 \text { vibrating upward, displacement increasing, acceleration also increasing, so D } }$ is correct;

Question 32

Question 32: As shown in Fig. (a), in the same medium, $\mathrm { S } _ { 1 } , \mathrm {~S} _ { 2 }$ two wave so...

As shown in Fig. (a), in the same medium, $\mathrm { S } _ { 1 } , \mathrm {~S} _ { 2 }$ two wave sources vibrate at the same time at $t = 0$, forming a wave with the same frequency. In the same medium, $\mathrm { S } _ { 1 } , \mathrm {~S} _ { 2 }$ two wave sources vibrate at the same time at $t = 0$, forming two columns of mechanical waves with the same frequency, and the vibration image of the wave source ${ } ^ { \mathrm { S } _ { 1 } }$ is shown in Fig. (b). diagram is shown in Fig. (c). The $P$ is a point in the medium, and the distances between the $P$ point and the wave source $\mathrm { S } _ { 1 } , \mathrm {~S} _ { 2 }$ are $7 \mathrm {~m} , 10 \mathrm {~m}$. The following statements are correct ( ) ![](/images/questions/phys-waves/image-020.jpg) ![](/images/questions/phys-waves/image-021.jpg) Figure (b) ![](/images/questions/phys-waves/image-022.jpg) Figure (c)

A. A. The vibration direction of the mass point $P$ is along the negative direction of the $y$ axis.
B. B. The amplitude of the mass $t = 1 \mathrm {~s}$ after $P$ is 1 cm.
C. C. $t = 1.25 \mathrm {~s}$ when the displacement of the mass $P$ is 0
D. D. In the time from $t = 0$ to $t = 2 \mathrm {~s}$, the distance traveled by the point $P$ is 112 cm.

Answer

Answer: D

Solution: A. Combining the image of the wave source ${ } ^ { S _ { 2 } }$ in ${ } ^ { t = 0.25 s }$ with the image of the wave in Fig. (c), it can be seen that at this time the vibration direction of the mass that has just begun to vibrate is along the negative direction of the ${ } ^ { y }$ axis, and the vibration direction of the mass and the wave source are the same, therefore, the source is ${ } ^ { S _ { 2 } }$, and the direction of vibration is along the negative direction of the ${ } ^ { y }$ axis. FORMULA_11]] is in the negative direction along the ${ } ^ { y }$ axis. According to the vibration image of wave source ${ } ^ { \mathrm { S } _ { 1 } }$ shown in Fig. (b), the vibration direction of wave source ${ } ^ { \mathrm { S } _ { 1 } }$ is upward. $$ P S _ { 1 } < P S _ { 2 } $$ is the wave source ${ } ^ { \mathrm { S } _ { 1 } }$. It can be seen that the vibration of wave source ${ } ^ { \mathrm { S } _ { 1 } }$ is first transmitted to ${ } ^ { P }$, so the vibration direction of ${ } ^ { P }$ is the same as that of wave source ${ } ^ { \mathrm { S } _ { 1 } }$, and it is the same as that of wave source ${ } ^ { \mathrm { S } _ { 1 } }$, and it is the same as that of wave source ${ } ^ { \mathrm { S } _ { 1 } }$ along the ${ } ^ { \mathrm { S } _ { 1 } }$. FORMULA_19]] axis, so A is wrong; B. From the figure BLOCK_FORMULA_1]] is the same vibration direction along ${ } ^ { y }$. Then the wave speed is $$ v = \frac { \lambda } { T } = \frac { 2 } { 0.2 } \mathrm {~m} / \mathrm { s } = 10 \mathrm {~m} / \mathrm { s } $$ $\mathrm { S } _ { 1 }$ and $\mathrm { S } _ { 2 }$ The time for the vibration to travel to $P$ is respectively \$\$\begin{aligned} \& t _ { 1 } = \frac { P S _ { 1 } } { v } = \frac { 7 } { 10 } \mathrm {~s} = 0.7 \mathrm {~s} <br> \& t _ { 2 } = \frac { P S _ { 2 } } { v } = \frac { 10 } { 10 } \mathrm {~s} = 1 \mathrm {~s} \end{aligned}$$ $P$ 点到两波源的波程差为 $$\Delta x = P S _ { 2 } - P S _ { 1 } = 10 \mathrm {~m} - 7 \mathrm {~m} = 3 \mathrm {~m}\$\$ The difference in wave range from point $P$ to the two sources is an odd multiple of half wavelength, and the vibration of the two sources is in opposite directions, so point $P$ is the point of vibration enhancement, and the vibration of the wave source ${ } ^ { \mathrm { S } }$ is transmitted to the ${ } ^ { 1 \mathrm {~s} }$ when the wave source ${ } ^ { 1 \mathrm {~s} }$ is just in the middle of the wave. The wave source ${ } ^ { 1 \mathrm {~s} }$ reaches point ${ } ^ { P }$ at ${ } ^ { P }$, so the amplitude of the mass point ${ } ^ { P }$ after ${ } ^ { 1 \mathrm {~s} }$ is 5 cm, so B is wrong; C. It can be seen that at ${ } ^ { t = 1.25 s }$, the wave source ${ } ^ { \mathrm { S } _ { 1 } }$ is at the point of ${ } ^ { t = 1.25 s }$. $$ t = 1.25 \mathrm {~s} - t _ { 1 } = 1.25 \mathrm {~s} - 0.7 \mathrm {~s} = 0.55 \mathrm {~s} $$ At ${ } ^ { P }$, the vibration condition of wave source $\mathrm { S } _ { 1 }$ at $$ t = 1.25 \mathrm {~s} - t _ { 1 } = 1.25 \mathrm {~s} - 0.7 \mathrm {~s} = 0.55 \mathrm {~s} $$ is transmitted to the point ${ } ^ { P }$, and at this time the wave source $\mathrm { S } _ { 1 }$ is located in the valley of the wave; and the wave source ${ } ^ { \mathrm { S } }$ at $$ t = 1.25 \mathrm {~s} - t _ { 2 } = 1.25 \mathrm {~s} - 1.0 \mathrm {~s} = 0.25 \mathrm {~s} $$ D. At $t = 1.25 \mathrm {~s}$, the particle $P$ is in the valley; D. In the time from $t = 0.7 \mathrm {~s}$ to $t = 1 \mathrm {~s}$, the distance traveled by particle $P$ is $$ s _ { 1 } = 6 A _ { 1 } = 12 \mathrm {~cm} $$ In the time from $t = 1 \mathrm {~s}$ to $t = 2 \mathrm {~s}$, the distance traveled by the particle $P$ is $$ s _ { 2 } = 5 \times 4 \left( A _ { 1 } + A _ { 2 } \right) = 100 \mathrm {~cm} $$ In the time from $t = 0$ to $t = 2 \mathrm {~s}$, the distance traveled by the point $P$ is 112 cm, so D is correct.

Question 33

Question 33: 38. An image of a simple harmonic transverse waveform at time $t = 0$ is shown in the figure. After ...

38. An image of a simple harmonic transverse waveform at time $t = 0$ is shown in the figure. After $\Delta t = 0.4 \mathrm {~s}$ time, exactly the first repetition of the waveform shown occurs. Based on the above information, the following options cannot be determined ( ) ![](/images/questions/phys-waves/image-023.jpg)

A. A. The magnitude of the speed of propagation of the wave
B. B. $\Delta t = 1.2 \mathrm {~s}$ The distance traveled by the point P in time.
C. C. $t = 0.6 \mathrm {~s}$ Velocity direction of the point P at the time
D. D. $t = 0.6 \mathrm {~s}$ Waveform of the moment

Answer

Answer: C

Solution: A. From the figure, we can see that the wavelength of the transverse wave is 8 m, from the meaning of the question can be obtained $$ \Delta t = T = 0.4 \mathrm {~s} $$ Then the magnitude of the speed of propagation of the wave is $$ v = \frac { \lambda } { T } = \frac { 8 } { 0.4 } \mathrm {~m} / \mathrm { s } = 20 \mathrm {~m} / \mathrm { s } $$ Therefore, A does not fit the question; B. According to the meaning of the question can be known, $\Delta t = 1.2 \mathrm {~s}$ time, the mass point vibration of the 3 cycles, the mass point P through the distance is $$ s = 3 \times 4 A = 120 \mathrm {~cm} $$ Therefore, B does not fit the question; C. Because the direction of wave propagation is not clear, it is not possible to determine the direction of the speed of point P. Therefore, C is consistent with the meaning of the question; D. When $t = 0.6 \mathrm {~s}$, the wave propagates $\frac { 3 } { 2 } T$, then the waveform is symmetric with the ${ } _ { t = 0 }$ moment about the $x$ axis, that is, the waveform at the $t = 0.6 \mathrm {~s}$ moment can be determined. The waveform at the $t = 0.6 \mathrm {~s}$ time can be determined.

Question 34

Question 34: 39 . The following statements about waves are correct

39 . The following statements about waves are correct

A. A. If there is a mechanical wave, there must be a mechanical vibration, and if there is a mechanical vibration, there must be a mechanical wave.
B. B. Transmission Enhancement Films on Camera Lenses as an Application of Light Polarization
C. C. We can tell that a distant planet is approaching us by the lowered frequency of light waves we receive on Earth from that planet.
D. D. The phenomenon of "hearing the sound before seeing the person" indicates that sound waves are more susceptible to significant diffraction than light waves.

Answer

Answer: D

Solution: A. There are mechanical waves must be mechanical vibration, mechanical vibration does not propagate the vibration of the medium, there will be no mechanical waves, so A error; B. The camera lens on the transmittance film utilized by the principle of light interference made, so B error; C. According to the Doppler effect can be known, received on Earth from a distant planet of the frequency of light waves become lower, you can judge that the planet is far away from us, so C error; D. The end of the first sound, because the sound wave wavelength is longer, easy to diffraction phenomenon; and the wavelength of the light wave is shorter, not easy to diffraction phenomenon, that is, the sound wave is more likely to diffraction than the light wave, so D error.

Question 35

Question 35: A simple harmonic transverse wave propagates in the negative direction of the $x$ axis, and the wave...

A simple harmonic transverse wave propagates in the negative direction of the $x$ axis, and the waveform at $t = 0$ is shown in the figure, and the wave propagates to $x = 1 \mathrm {~m}$ at that moment. The coordinates of the equilibrium positions of the $P , Q$ two points at $( - 1,0 ) \quad ( - 7,0 )$ are $( - 1,0 ) \quad ( - 7,0 )$. If it is known that a wave trough occurs for the second time at ${ } ^ { t = 0.6 \mathrm {~s} }$ at the point of mass $P$, then the following statement is correct ![](/images/questions/phys-waves/image-024.jpg) High School Physics Assignment, October 29, 2025

A. A. The wave has a wavelength of 4 m
B. B. The wave travels at $\frac { 10 } { 3 } \mathrm {~m} / \mathrm { s }$
C. C. The origin direction of the vibration source is along the negative direction of the $y$ axis.
D. D. When the prime $Q$ is at the crest of the wave, the prime $P$ is at the crest of the wave

Answer

Answer: A

Simple Harmonic Motion and Mechanical Waves

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