# Simple Harmonic Motion

1. May 10, 2009

### jjlandis

1. The problem statement, all variables and given/known data

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 2.13 m and frequency 0.385 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.

2. Relevant equations

V(ox) = -$$\varpi$$*A*sin$$\phi$$

3. The attempt at a solution

I notice it is a simple harmonic motion question so I am using the equation relative to SRH velocities. Since it is a vertical motion, the gravity and mass play a role in the velocity. The angle I understand to be zero degrees?

I first found angular frequency by taking 2*pi*frequency. Then multiplied it by amplitude to get 5.15. The answer is 3.18. I know there is something small i'm missing, probably the mass factor, but i've been stuck for a while and could use some help....

2. May 10, 2009

### diazona

What angle?

Do you understand what that equation means?

3. May 10, 2009

### jjlandis

The equation assumes t = 0 and v(ox) = v(x). I figured the initial velocity would be the instant the rider left the bull. the angle is the phase angle, and without an x direction, I didn't think there would be any angle. and since amplitude and frequency were given, that left me only a V(ox) variable to calculate.

I also tried to use:

Vmax = $$\varpi*Amplitude$$, from which I was initially under the impression was the correct way to do it but I am not returned close to the correct answer

4. May 10, 2009

### diazona

Sounds like you're confused about what a "phase angle" is. The real equation is
$$v = -A \omega\,\mathrm{sin}(\omega t)$$
The $$\phi$$ in your equation is equal to $$\omega t$$ - it increases with time. It's not a real angle in space.

Now, think about what happens when you leave the bull flying upward: it means that the acceleration of gravity is not strong enough to hold you down to the bull. In other words, the bull would be accelerating downwards faster than the acceleration of gravity. Figure out how fast the bull is moving when that condition is met, and I think you'll have your answer.

5. May 10, 2009

### jjlandis

I think I am confused about "phase angles" as well.

Figure out how fast the bull is moving once it exceeds the acceleration of gravity?
Does that mean I am still using the SHM veloctiy and acceleration equations?