I don t have an idea for this question of physics waves, can you help analyze it, thank you 5

This is actually very simple, and you can see that the period of this wave is 4s, and the key is the distance between the two points of ab, so how to determine his distance? At 0s, point A is at the origin and is moving downward, while point B is at the highest peak and is also moving downward. Then it is clear that they are (n+1 4) wavelengths apart, then it can be calculated.

It is possible to know that the wavelength of n is not in the required range. When n=2, the wavelength is 4, and the wave velocity is 1

a, b spacing is a quarter wavelength plus an integer wavelength, that is, ab = 9 m = (1 4 + n), = 9 (1 4 + n), according to the title: 3< = < = 7, substituted into the above equation to get: < = n<= Therefore, n can take 1 or 2, then can take meters or 4 meters, and the period is 4 seconds, so the wave speed can be meter seconds, or 1 meter second.

When t=0, a is in the equilibrium position and b is at the positive maximum displacement. Since the wave propagates to the right, the length between A and B may be 1 4 wavelengths or 5 4 wavelengths, etc., and the relationship is as follows (1 4+N) = 9 m. Since 3m 7m, n=2, =4m, the condition is satisfied, v=1ms.

Take t=2 to study, A is in the equilibrium position and the movement trend is upward, and B is in the trough and the movement trend is upward.

Assuming that Figure B is a Y-X plot, then A is at T=1s and B is at T=2S.

It is easy to know that a and b are (1 4+n)t, and t=4s is known from the diagram because (1 4+n)t is 9m, and 3m is 7m

From this we can solve n=1, 9 5=wave velocity=

A is in equilibrium and the movement is trending upwards, B is in the trough and the movement is trending upwards.

Assuming that Figure B is a Y-X plot, then A is at T=1s and B is at T=2S.

It is easy to know that a and b are (1 4+n)t, and t=4s is known from the diagram because (1 4+n)t is 9m, and 3m is 7m

This solves n=2, 4m

Wave velocity = 4 4 1 m s

The picture on the far left was drawn by yourself, right? is wrong.

y(b)=a cos(ωt) y(a)=a cos(ωt+9ω/u)=-a sin(ωt)=a cos(ωt+π/2+2kπ)

9 u= 2+2k 3<= <=7 so 1<=u<=7 3 t=4s u=1m s

u*t=4m

The wavelength is 4 meters, and the wave speed is 1 meter per second.

I'm just here for that score, no, I can't help you.

I can't see it clearly, can you give it a little clearer?

There are two ways to explain that b vibrates at least 3 4 cycles later than a, first saying why 3 4 cycles, and then saying why at least.

One: Analyze from zero moment. Because the wave is transmitted from A to B, it is necessary to analyze the time when B obtains the state of A, and look at the 0 moment of the vibration ** image, B is in the equilibrium position and moves upward, and A is in the trough of the wave, so B needs 3 4 cycles.

And due to the periodicity, B can also obtain the state of A after n+3 4 cycles. Therefore b vibrates at least 3 4 cycles later than a.

Two: start from the time. Since the wave is transmitted from A to B, so first find A (this can find a special point, it can be a trough or a peak, but pay attention to wait for B to find the same special state), such as at the end of 2 seconds, A is at the crest, and the back (note that it is the back, because if it is from A to B, so B must be backward in time) B only appears in 5 seconds, so B is 3 seconds behind, and the period is 4 seconds, that is, B is behind 3 4 cycles, and finally with the same reason periodicity, B may be in 9 seconds, 13 seconds, 17 seconds.

There is a peak, so B is behind N+3 for 4 periods.

The force of the object is balanced, then the elastic force of spring B to the object is upwards of 2 3g, and spring A should also provide a tensile force of 1 3 upwards.

Let spring B return to the distance of 1 3 and lift 1 3 upwards, at this time spring A will have to go up 1 3 distance, so it is a total of 2 3 l.

Both springs should be elongated. First, analyze A and the object: the upward pull force, the downward pull of the spring, and the gravitational force. Spring B is elongated by 1 3L, and Spring A is also elongated by 1 3L when it is subjected to an upward pull force FSo for 2 3 l

mg=fl, when B bears the weight of 2 3 things, the compression amount is 2 3L, then the object needs to rise by 1 3L, and the weight of the other 1 3 objects is borne by A, and the elongation is 1 3L, so the two parts add up to 2 3L

From the waveform of the figure, we know that the direction of the source of P is upward, and the direction of the source of Q is downward, and the wave of P arrives first, so A is wrong. The wave velocity is the same in the same propagation medium, so the p source is closer to the vibration system, so b is wrong. It is explained that the vibration system has resonated, so the frequency of the Q source is also 2Hz, so the wave velocity of the Q source is 4m s, and C is correct.

When three objects are stacked, when the horizontal force acting on object B is f=2n, and all three objects are stationary, then the friction between objects A and B is 0 N, the friction between B and C is 2 N, and the friction between C and the ground is 2 N

I wonder why there is friction between C and the ground, is it under tension?

Because the three objects are at rest and belong to the static friction force, as long as the two objects have a relative tendency to motion, there is static friction and the magnitude is equal to the external force, and the direction is opposite. Therefore, the magnitude of the friction between b and c is 2n, and the friction force is in pairs, and the friction force of b by 2n is opposite to the horizontal force f, and the friction force of c by 2n is the same as that of the horizontal force f, so c also has a relative motion trend relative to the ground, that is, there is also friction between c and the ground.

0，2，2

The tensile force experienced by C is actually the friction force of the right that B acts on C horizontally to the right, so the friction between C and the ground is horizontally to the left.

2 2 2

C is rubbed by B, there must be a force balanced with it, that is, the ground gives it, C is not balanced if it is not rubbed by the ground, this C is not only relative to B, but also has friction relative to the ground.

0 2 2 Because AB has a frictional force to the right towards C, and C does not move, it must be balanced by force, and only the ground can give the force, so it is inferred from the equilibrium principle that the ground has a leftward friction force to C.

2 ,2,2

There is static friction between c and the ground, otherwise the three objects would not all be at rest.

The answers are 0 2 2 respectively, the action of force is mutual, B gives C a pulling force and C gives B a reaction force.

It's not the pulling force, it's the frictional force.