- #1
NotMrX
- 97
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A wire could be shaped like a sinusoidal function and then we could say the bead moves harmonically.
The shape of the wire, so that bead occilates around X = 0.
Y = -50*cos(10X)
If we ignore friction and give it a small dispalcement then it is possible to find angular frequency.
However, when I applied my method to a varation of this problem in a textbook I got a wrong answer.
Part I:
Here was my attempt where Z is the angle from the center of the circle that the bottom of the sinudosoidal function fits on:
S = (r)*(Z) = (50)*(Zmax sin wt)
V= 50*Zmax*w*coswt
V(max) = 50*Zmax*w
Part II:
Energy at the lowest place
E = .5 m (Vmax)^2 - 50mg
Energy at the highest place
E = -50mg*cos(Zmax)
Setting the energies equal:
.5 m (Vmax)^2 - 50g =-50mg*cos(Zmax)
Solving for the velocity:
(Vmax)^2 = 100g*[1-cos(Zmax)]
Part III: combing part I & II
(Vmax)^2 = 100g*cos(Zmax)
(50*Zmax*w)^2 = 100g*[1-cos(Zmax)]
library logged me off, i will finish later
The shape of the wire, so that bead occilates around X = 0.
Y = -50*cos(10X)
If we ignore friction and give it a small dispalcement then it is possible to find angular frequency.
However, when I applied my method to a varation of this problem in a textbook I got a wrong answer.
Part I:
Here was my attempt where Z is the angle from the center of the circle that the bottom of the sinudosoidal function fits on:
S = (r)*(Z) = (50)*(Zmax sin wt)
V= 50*Zmax*w*coswt
V(max) = 50*Zmax*w
Part II:
Energy at the lowest place
E = .5 m (Vmax)^2 - 50mg
Energy at the highest place
E = -50mg*cos(Zmax)
Setting the energies equal:
.5 m (Vmax)^2 - 50g =-50mg*cos(Zmax)
Solving for the velocity:
(Vmax)^2 = 100g*[1-cos(Zmax)]
Part III: combing part I & II
(Vmax)^2 = 100g*cos(Zmax)
(50*Zmax*w)^2 = 100g*[1-cos(Zmax)]
library logged me off, i will finish later