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Using Simple Harmonic motion and conservation of motion to find maximum velocity

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Fromaginator
#1
Nov29-09, 01:52 AM
P: 8
The Question


Relevant equations
KE=0.5*m*v^2
T=2*pi*sqrt(m/k)
EE=0.5*k*x^2
KEi+UEi=KEf+UEf
I think that's all of them

attempt at a solution
I was thinking about just using the conservation of energy at the the equilibrium point, as the kinetic energy would be at a max there and the potential energy would be at its minimum, Also the velocity would be at its max while the acceleration would be zero. However, one of the many problems I've run into on this question is that I'm unsure if the average of the two x values will give me the equilibrium point (if it did then it would be 5mm) and I don't feel comfrontable moving on having possibly made a false assumption.
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Fromaginator
#2
Nov29-09, 01:59 AM
P: 8
I got the answer :) but for others what I did was
from the graph u can see that potential at highest point is 5J and at lowest point is 1J

mgh+ 1/2mU^2 = mgH + 1/2 mV^2 in this U = initial velocity = 0 and V = max velocity at lowest point and mgh is initial potential energy and mgH is potential energy at lowest point.
this gives
5 + 0 = 1 + (1/2 x (1.7/1000) x V^2) [ 1.7 is divided by 1000 to convert grams into kg]
V^2 = 4705.88
V = sqrt(4705.88)
V = 68.59 m/s
AakashR
#3
Dec6-09, 08:44 PM
P: 8
Remember one thing- All the oscillations have two things in common
I. the oscillation takes place about an equilibrium position and,
II. the motion is periodic.

These are the only two conditions that are to be fulfilled for a motion to be oscillatory motion.

Fromaginator
#4
Dec7-09, 01:46 PM
P: 8
Using Simple Harmonic motion and conservation of motion to find maximum velocity

Thanks I'll keep that in mind in similar problems, my finals coming up soon it may come in handy


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