Determining the period of a periodic motion

  • #1
insertnamehere
50
0
Hi. I am currently working on a problem involving a ball being dropped from a hieght of 4m making a perfectly elastic collision with the ground. Assuming no mechanical energy is lost due to air resistance, I must find the period of the motion.
I used the equation
V(final) = V(initial) - gt
where V(initial)=0
therefore V(final)= -9.8t

and V^2(final) = V^2(initial) - 2gh
h= -4m
V(final)= -9.8t
so i replaced these values in the equation, and i find t= 0.9035 seconds
To find T, I simply multiplied 0.9035 sec by 2. However, my answer does not correspond that what it says in the book, and now I'm lost! Can someone help me by telling me if my method is right or wrong? I would really appreciate it!
Thank you!
 

Answers and Replies

  • #2
insertnamehere
50
0
And also, why is this type of motion NOT Simple Harmonic?
 
  • #3
whozum
2,221
1
It is simple harmonic motion if the collisions are completely elastic and no air resistance/friction is present.

It takes 1.807s for the ball to fall, hit the ground, and go back to its original height. What does the book say?
 
  • #4
lightgrav
Homework Helper
1,248
30
1) it would have been more straightforward to use
x = x_0 + v_0 *t + .5 g t^2 => 4m = .5(9.8m/ss) t^2

2) yes, the time Period T = 1.807s . Did they give frequency?

3) "Harmonic" means location function x(t-t_0) can be written
as a Sum of sine waves with (a) w_i = n * w_0
"Simple Harmonic" means that there's only one term:
x(t) = A sin(wt) (if you set your starting time right, so no phase).
Do you think this ball's motion is Simple Harmonic?
 

Suggested for: Determining the period of a periodic motion

  • Last Post
Replies
8
Views
29
Replies
4
Views
256
  • Last Post
Replies
4
Views
300
  • Last Post
Replies
3
Views
364
Replies
1
Views
361
Replies
5
Views
376
  • Last Post
Replies
22
Views
1K
Replies
3
Views
116
  • Last Post
Replies
3
Views
616
Replies
6
Views
306
Top