Put a spring on the ground. Now put a small ball on the spring lightly. At this time, the spring will compress. When the elastic force equals the gravity of the small ball, will the small ball still press down? After reaching the lowest point, will the point where the elastic force equals the gravity be taken as the equilibrium position to make a simple harmonic motion? Why not do simple harmonic motion?

analysis:
If the air resistance is not considered, and the spring length is long enough (the spring has enough space to compress), then in the ideal case (the spring is absolutely vertical, and the ball will not fall from the edge), the ball will do simple harmonic motion on the spring
When the spring is in the original long state, the small ball starts to put on the top of the spring (without initial velocity), and the small ball and the spring form a "spring vibrator". At this time, when the spring force is 0, the small ball will accelerate to drop and compress the spring. When the spring force is equal to the gravity of the small ball (the resultant force is 0), the speed of the small ball is the maximum (this position is the equilibrium position); because the small ball has inertia, it will continue to move downward, Make the spring force (upward) greater than the gravity of the ball, so the ball will continue to slow down until the speed is 0 (this position is the maximum displacement)
Because the elastic force at the lowest point (maximum displacement) is greater than gravity, the ball will accelerate upward and reach the maximum speed at the equilibrium position; similarly, when it continues to move upward, the elastic force is less than gravity, and the ball will decelerate until the speed at the highest point (corresponding to the original length of the spring) is 0
In the above process, the ball will never leave the spring, and the spring oscillator composed of the ball and the spring does do simple harmonic motion

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