Simple harmonic motion of a ball drop

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Homework Help Overview

The discussion revolves around the motion of a ball dropped from a height of 4 meters, which makes a perfectly elastic collision with the ground. The participants explore concepts related to periodic motion, the determination of the period, and whether the motion qualifies as simple harmonic motion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the periodic nature of the motion, particularly regarding the direction of the force acting on the ball. There is also discussion about the calculations for the period of the motion and whether momentum is conserved during the collision with the ground.

Discussion Status

Some participants have provided insights into the periodic nature of the motion and the conservation of energy during the elastic collision. There is ongoing exploration of the definitions and characteristics of simple harmonic motion, with differing opinions on whether the motion fits that classification.

Contextual Notes

Participants are considering the implications of elastic collisions and the nature of forces acting during different phases of the ball's motion. The discussion includes assumptions about energy conservation and the behavior of the ball after impact.

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)
[tex]t=\sqrt{\frac{2d}{a}}[/tex]
[tex]t=\sqrt{\frac{2*4m}{9.8}}[/tex]
[tex]t=0.9035s[/tex]

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 [itex]\vec{F}=-k\vec{r}[/itex] always implies a harmonic oscillation with period [itex]T=2\pi\sqrt{\frac{m}{k}}[/itex].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 [tex]0< t< [\sqrt{\frac{8}{g}}[/tex], then [tex]-\frac{g}{2}(t-\sqrt{\frac{8}{g}})^2[/tex] for [tex]\sqrt{\frac{8}{g}}< t< 2\sqrt{\frac{8}{g}}[/tex].
 

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