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## Homework Statement

While visiting friends at Cal State Chico, you pay a visit to the Crazy Horse Saloon. This fine establishment features a 200- kg mechanical bucking bull that has a mechanism that makes it move vertically in simple harmonic motion. Whether the “bull” has a rider or not, it moves with the same amplitude 1.61 m and frequency 0.474 Hz. Being from Texas you decide to ride it the “macho” way by NOT holding on. To no ones surprise you go flying out of the saddle. While waiting for your bruises and pride to heal, you decide to calculate how fast upward you were moving when you left the saddle.

Give your answer in m/s to the second decimal place.

## Homework Equations

1. [tex]1/2\,m{v}^{2}+1/2\,k{x}^{2}=1/2\,k{A}^{2}[/tex]

2. [tex]v=\sqrt {{\frac {k}{m}}}\sqrt {{A}^{2}-{x}^{2}}[/tex]

3. [tex]v=\omega\,A[/tex]

4. [tex]\omega=2\,\pi\,f[/tex]

## The Attempt at a Solution

I had several attempts, all of which get me in the right ballpark, but not the correct answer, which is 3.49 m/s.

1) Using the third and fourth equations:

[tex]v=2\,\pi\,fA[/tex]

= 4.79.

After thinking about it some more, this equation won't work because this finds the maximum velocity while still connected to the spring.

2) I attempted to use the second equation, but again, this would mean I was still attached to the bull, and thus x = A, making v = 0.

3) I reread the problem and saw that it said AFTER I left the bull, what was my speed. So I thought, conservation of energy:

[tex]1/2\,k{A}^{2}=1/2\,m{v}^{2}+mgh[/tex], with h = A.

then solving for v:

[tex]v=A\sqrt {{\frac {k}{m}}-2\,gA}[/tex], but I still needed to get rid of the k, so I thought:

[tex]\omega=\sqrt {{\frac {k}{m}}}[/tex], but this is where I am almost certain that my leap of faith isn't true; however, I continued, using the fourth equation to substitute for [tex]\omega[/tex], and got -.82 m/s...

So, I'm stuck. I think I'm on the right track with conservation of energy, but I don't think I can create [tex]\omega[/tex] into that equation.