Bead on a Wire and Harmonic Motion

[NotMrX]

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

 

[NotMrX]

Nevermind I figured it out. I made mistake in my method before.

s = r Z

r = 1/k where k is the curvature.

 

[kyle8921]

I would like to provide some clarifications and suggestions for the above content. Firstly, the concept of a bead on a wire and harmonic motion is a well-known phenomenon in physics. When a bead is attached to a wire and given a small displacement, it will oscillate back and forth around a point of equilibrium (in this case, X=0) due to the restoring force of the wire. This oscillation can be described by a sinusoidal function, as shown in the equation Y=-50*cos(10X).

However, it is important to note that the wire itself does not necessarily need to be shaped like a sinusoidal function. The shape of the wire only affects the amplitude and frequency of the oscillation, not the nature of the motion itself. In other words, a straight wire can also produce harmonic motion if the bead is given a small displacement.

In Part I, the equation S=(r)*(Z) is not entirely accurate. The displacement of the bead from the center of the circle (Z) should be multiplied by the angular frequency (w) to get the displacement (S) at any given time. Additionally, the equation for velocity (V) should include the angular frequency (w) as well, as it affects the rate of change of displacement. Therefore, the correct equations would be S=(r)*Z*sin(wt) and V=(r)*w*cos(wt).

In Part II, the equation for energy at the highest place is not entirely accurate. The correct equation should be E=0.5*m*(Vmax)^2+50mg, as the bead has a non-zero velocity at the highest point of its motion. Additionally, the equation for energy at the lowest place should include the potential energy due to the displacement of the bead from the center of the circle, which would be 0.5*m*(w*S)^2. Therefore, the correct equations would be E=0.5*m*(Vmax)^2+50mg and E=0.5*m*(w*S)^2-50mg.

In Part III, the equations for energy at the highest and lowest points should be set equal to each other, not to -50mg. This is because the total energy of the system (bead and wire) remains constant throughout the motion. Therefore, the correct equation would be 0.5*m*(Vmax)^2+50mg=0.5*m*(w*S)^