Simple harmonic motion of a ball drop

In summary, a ball dropped from a height of 4m and makes a perfectly elastic collision with the ground will exhibit periodic motion due to the conservation of energy. The period of the motion is equal to the time it takes for the ball to fall and bounce back to its initial position, which can be calculated using the equation t=\sqrt{\frac{2d}{a}}. This motion is not simple harmonic as the force acting on the ball is not always directed towards the equilibrium position. The net force during the collision with the ground is upward, allowing the ball to return to its initial height and repeat the same motion.
  • #1
UrbanXrisis
1,196
1
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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  • #2
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.
 
  • #3
what is the period? Did I do it correctly? and why is it periodic?
 
  • #4
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.
 
  • #5
so momentum is not lost after hitting the Earth making the ball bounce back to its original place right?
 
  • #6
U can compute the variation in momentum and see that it is not negative...

Daniel.
 
  • #7
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].
 

What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth along a linear path, with its displacement from its starting point being directly proportional to the force acting on it.

What is the equation for the period of a simple harmonic motion?

The period of a simple harmonic motion can be calculated using the equation T = 2π√(m/k), where T is the period in seconds, m is the mass of the object in kilograms, and k is the spring constant in newtons per meter (N/m).

How is the amplitude of a ball drop related to its potential energy?

The amplitude of a ball drop, or the maximum displacement of the ball from its equilibrium position, is directly proportional to the potential energy of the ball. As the amplitude increases, so does the potential energy, and vice versa.

What factors affect the frequency of a ball drop's simple harmonic motion?

The frequency of a ball drop's simple harmonic motion is affected by the mass of the ball, the spring constant of the surface it is dropped on, and the amplitude of the oscillation.

Can a ball's simple harmonic motion continue indefinitely?

In theory, a ball's simple harmonic motion can continue indefinitely as long as there is no external force acting on it. However, in real life, there will always be some friction or air resistance present, causing the ball to eventually come to a stop.

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