Energy and Simple Harmonic Motion - Part II.

  • #1
A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall whle the other end is connected to a movable object. The spring and object are compressed by 0.064 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 11.5 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.037 m relative to its unstrained length?

... I tried using KEo + PEo + SPEo = KEf + PEf + SPEf ... but, I can't get anywhere ...

... I found out that the frequency is 18.05hz and it's relative time is .056 seconds - and that's where I'm stuck!
 

Answers and Replies

  • #2
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7
Use Hooke's law and conservation of energy.
 
  • #3
Do I set them equal to each other??

... I don't understand the .64 meter compression part ... does that equal "X" in Hooke's Law and in ????
 
  • #4
Hint -- Can you find the value of the spring constant?

BTW -- you are right in that the compression WILL be X in hooke's Law.
 
  • #5
I figured it out ...

A = 0.064m (given in the problem)
omega = 11.5 (rad/s) - given in the problem
x = 0.037m (given in the problem)


V(max) = A(omega)
V(max) = 0.064(11.5)
V(max) = .736(m/s)

V = V(max) * [the square root of the quantity of (1- {[x^2]/[A^2]})]
V = .736 * [the square root of the quantity of (1- {[0.037^2]/[0.064^2]})]

V = 0.601 (m/s)
 

Related Threads on Energy and Simple Harmonic Motion - Part II.

  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
5
Views
8K
  • Last Post
Replies
1
Views
2K
Replies
16
Views
2K
Replies
4
Views
13K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
17K
Replies
1
Views
5K
Top