Simple harmonic motion of a ball drop

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SUMMARY

The discussion centers on the analysis of a ball dropped from a height of 4 meters, demonstrating that its motion is periodic but not simple harmonic. The period of the motion is calculated to be 1.8075 seconds, derived from the formula t = √(2d/a), where d is the drop height and a is the acceleration due to gravity (9.8 m/s²). Although the motion is periodic due to the elastic collision with the ground conserving energy, it does not qualify as simple harmonic motion since the acceleration is not proportional to the displacement from an equilibrium position.

PREREQUISITES
  • Understanding of basic physics concepts such as periodic motion and elastic collisions.
  • Familiarity with kinematic equations, particularly t = √(2d/a).
  • Knowledge of gravitational acceleration (9.8 m/s²) and its role in free fall.
  • Concept of simple harmonic motion and its mathematical representation.
NEXT STEPS
  • Study the principles of elastic collisions and energy conservation in physics.
  • Learn about the characteristics and equations of simple harmonic motion.
  • Explore the mathematical modeling of periodic motion using differential equations.
  • Investigate the effects of air resistance on the motion of falling objects.
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Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of periodic motion and elastic collisions.

UrbanXrisis
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a ball is dropped from a heright of 4m and makes a perfectly elastic collision with the ground. Assuming that no energy is lost due to air resistance,

a) show that the motion is periodic

b) determine the period of the motion

c) is the motion simple harmonic?

a) I know that periodic motion mneas that the force is always directed towards the equilibrium position, making a back and forth motion. However, the force is always downwards. Does that mean this motion is not periodic? Not quite sure how to answer this.

b)
t=\sqrt{\frac{2d}{a}}
t=\sqrt{\frac{2*4m}{9.8}}
t=0.9035s

This is the time it takes for the ball to fall, does it lose momentum since it transferes it to the ground? That would make the initial velocity after hitting the ground smaller. If momentum is not lost, then the period would be 1.8075s.

c) Simple harmonic motion is when an object's acceleration is porportional to its displacement from some equilibrium position and is oppositely directed. The acceleration is always downwards so there is no harmonic motion?
 
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The motion is periodic,but not harmonical.A force of type \vec{F}=-k\vec{r} always implies a harmonic oscillation with period T=2\pi\sqrt{\frac{m}{k}}.In your case,the force doesn't have that character (it does,but only when the ball is climbing).

Daniel.
 
what is the period? Did I do it correctly? and why is it periodic?
 
Because the dynamics is "symmetric" in time.It's like a wheel which rotates with constant angular velocity.The points of the whell move along circles and that movement is periodic:after a certain amount of time,they return to their initial position.So does the ball...

The period u calculated is fine.

Daniel.
 
so momentum is not lost after hitting the Earth making the ball bounce back to its original place right?
 
U can compute the variation in momentum and see that it is not negative...

Daniel.
 
Let me point out that the force is not "always downward". The net force during the collision with the ground is upward.

Periodic just means that it repeats in exactly the same way.
Since the collision with the ground is elastic (that's pretty hard ground!) total energy is conserved. Since the kinetic energy is 0 at the highest point, the ball returns to the same height. Then everything happens again in exactly the same way: periodic motion. The period is the time the ball takes to hit the ground, then bounce back to the same point. Even simpler, the ball will take the same time to bounce back up to the initial point as it took to drop to the ground- calculate that (you did) and double to find the period.

Simple harmonic motion is sinusoidal- the height h is a sine function of time. That is not true here: h(t)= -(g/2) t2+ 4 for 0< t< [\sqrt{\frac{8}{g}}, then -\frac{g}{2}(t-\sqrt{\frac{8}{g}})^2 for \sqrt{\frac{8}{g}}< t< 2\sqrt{\frac{8}{g}}.
 

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