# Simple harmonic motion of a ball drop

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?

dextercioby
Homework Helper
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?

dextercioby
Homework Helper
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?

dextercioby
Homework Helper
U can compute the variation in momentum and see that it is not negative...

Daniel.

HallsofIvy
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 $$\sqrt{\frac{8}{g}}< t< 2\sqrt{\frac{8}{g}}$$.