Determining speed of an oscillating object

  • #1

Homework Statement



(Forgive me if I use certain terms wrong; I'm learning all of this in French)

A branch at the top of a tree is swinging back and forth with a simple harmonic motion. Its amplitude is 0.80m. Its maximum speed at the point of equilibrium is 1.5m/s. What is the speed of the branch at 0.60m?

The answer according to the back of the book is 0.75m/s but I need to know how to get there.

Homework Equations



Em = Em'
Ep + Ec = Ep' + Ec'
Ep = kx2/2
Ec = mv2/2

The Attempt at a Solution



I've made many attempts at this, all completely wrong. I always end up trying to factorize (my teacher told us that the answer involves factorizing) but it goes absolutely haywire. It would take too long to type out even one attempt.

Em = Em'
Epmax = Ecmax
kxmax2/2 = mvmax2/2 (the 1/2 cancels for each formula)
vmax/xmax = √k/m (which should be constant)
(1,5m/s)2/(0.80m)2 = √k/m
1.875(not sure what unit to put here) = √k/m ∴ k/m = 3.516

kx2 + mv2 = mvmax2 (again the 1/2 cancels)
mvmax2 - mv2 = kx2
m(vmax2 - v2) = kx2
vmax2 - v2 = kx2/m
k/m = (vmax2 - v2)/x2

If the k/m stays constant, which I believe it should, it should be an easy substitution to find the v I'm looking for.

After manipulating the formula I got, I'm left with

v = √-(k/m - vmax2/x2)(x2)

After substituting, I get 0.99m/s.
 
Last edited:

Answers and Replies

  • #2
3,816
92
Start by finding out the angular frequency. :wink:

You are given maximum velocity and amplitude, can you find the angular frequency?
 
  • #3
I don't know what angular frequency is :S We never learned anything like that. (Physics 20)
 
  • #4
1,065
10
After substituting, I get 0.99m/s.
........
I get the same answer

v=VmaxSin(ArcCos(6/8))
V=1.5 X sin(41.41°)
v=0.99 m/s
 
Last edited:
  • #5
I get the same answer

v=VmaxSin(ArcCos(6/8))
V=1.5 X sin(41.41°)
v=0.99 m/s


Nope, I definitely haven't learned what you just did.

If it helps, I think I'm supposed to assume it's moving in a straight line like a spring oscillating back and forth. The branch should be moving in a partial circular path, but those calculations would be more advanced than what I've been taught.
 
Last edited:
  • #6
1,065
10
Ok it's more complicated than SHM. For SHM our answer is correct.
 
Last edited:

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