Derivation steps of velocity and displacement formula A = vt-v0 / T and S = average vt How to write the steps of deriving VT = at and S = at square · 1 / 2? Thank you

Derivation steps of velocity and displacement formula A = vt-v0 / T and S = average vt How to write the steps of deriving VT = at and S = at square · 1 / 2? Thank you

From a = (vt-v0) / T and S = V (average) t, it is very simple to deduce the velocity formula of uniform acceleration linear motion with zero initial velocity: from a = (vt-v0) / T, VT = V0 + at, because V0 = 0, VT = at deduces the displacement formula of uniform acceleration linear motion with zero initial velocity: because the average velocity of uniform acceleration linear motion V (average) = (V0 + VT) / 2V

Solution of displacement formula of simple harmonic motion
If we only know the definition of simple harmonic motion, that is, when an object is in simple harmonic motion, the force on the object is proportional to the displacement, without other conditions, how to deduce the displacement formula of simple harmonic motion?

I don't want to say more. It's right upstairs

It is proved that the vibration of spring oscillator in vertical plane is simple harmonic motion

Stress analysis, Hooke's law and Newton's second law, second order linear differential equation with constant coefficients, reduced order method for solving displacement function

When a vertical spring vibrator moves harmoniously in the vertical plane, the stiffness coefficient of the spring is 100N / m, the gravity of the vibrator is 5N, and the maximum elongation of the spring is 15cm, what is the maximum restoring force of the spring vibrator? When the vibrator passes through the highest point, what is the size of the spring?

The maximum restoring force of spring oscillator is Fmax = KX mg = 100 * 0.15-5 = 10N
Elongation of 5N weight hanging by spring: x = f / k = 5 / 100 = 0.05m = 5cm
The amplitude is 15cm-5cm = 10cm
When the vibrator passes through the highest point, the spring force is f = KX = 100 (10-5) * 10 ^ - 2 = 5N

Is the vertical motion of spring oscillator simple harmonic motion?
Why?

Yes
First of all, spring balance is just started:
mg=kx；
When pulling down:
F resultant force = mg-k (x + X ')
Combine the above two forms
F resultant force = - KX '

Why is the vibration of a spring oscillator a simple harmonic motion,

Recovery force F = - kx

When t = 0, one end of the rope starts to make a simple harmonic motion with a period of T. after time t (3T / 4 < T < T), a point on the rope is located at the maximum displacement above the equilibrium position. Then at 2T, the point is located at ()
A. Up and up B. up and down
C. Down and up D. down and down

After time t (3T / 4 ＜ t ＜ T), a point on the rope is located at the maximum displacement above the equilibrium position. At 2T, the particle vibrates again at time t. We can analyze the position and velocity direction of the particle according to the relationship between time and period
A period can be divided into four T 4, then at 2T, the point is above the equilibrium position and moves upward. Therefore, a is correct and BCD is wrong. If there is something incomprehensible, it can be explained in detail

If t '= t / 2, then the length of the spring at t is equal to that at t + T'?
This sentence is wrong. The teacher said that the shape variables must be equal. I don't understand the difference between shape variables and length in the model of spring oscillator?

It's not necessarily the same length
For example, take any point in the first quarter of a cycle, after t / 2, in the middle of T ~ 3t / 2, it is obvious that one is in the state of elongation, and the other is in the state of compression, the length is not equal, but the shape variables are equal. You are very powerful. You are very entangled with this problem, so that you can understand the knowledge, come on!

The vibration period of a spring oscillator with simple harmonic motion in the horizontal direction is 0.4s. When the oscillator starts to move to the right from the equilibrium position, the motion of the oscillator is ()
A. Decelerating to the left B. accelerating to the right C. acceleration decreasing D. kinetic energy decreasing

If the period of the oscillator is 0.4s, then the quarter period is 0.1s, and 0.05s is less than one quarter period; a. the oscillator starts to move to the right from the balance position, and does not reach the maximum displacement on the right at 0.05s, so the oscillator is decelerating at the maximum displacement on the right, so a and B are wrong; C. because the oscillator is moving to the maximum displacement on the right

When a spring oscillator makes a simple harmonic motion, it is calculated from the equilibrium position O. after 3S, the oscillator passes through point P for the first time, after 2S, and after the second time, can the particle pass through point P for the third time 10 / 3S?
I worked out two solutions, but the answer was not the 10 / 3S,

It's 10 / 3S. You're right