Simple Harmonic Motion of mechanical bull

[factor]

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 2.36 m and frequency 0.338 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


I would offer a proposed solution to this problem, but I'm honestly not even sure where to begin. Any help in pointing me in just the general direction would be greatly appreciated. I imagine what I'm missing is something really natural and intuitive. Some stupid, small mistake.

 

[radou]

I hope you'll find this link useful: http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html" .

 

[factor]

Well the problem is that I don't see a relation to mass on a spring as it says the bull moves the same regardless of the rider on it. And I tried finding the location where the rider would actually leave the saddle, I found that it of course is at the top of the bull's range of movement, at least that's what I would figure as the magnitude of the acceleration of the bull is always less than gravity meaning gravity keeps him on the seat until the direction of the bull's velocity changes. However, I'm just not sure how I would go about calculating his velocity at that point. Energy conservation didn't seem to work, but then again I may still be missing something simple in that case. Thanks.

 

[OlderDan]

Well the problem is that I don't see a relation to mass on a spring as it says the bull moves the same regardless of the rider on it. And I tried finding the location where the rider would actually leave the saddle, I found that it of course is at the top of the bull's range of movement, at least that's what I would figure as the magnitude of the acceleration of the bull is always less than gravity meaning gravity keeps him on the seat until the direction of the bull's velocity changes. However, I'm just not sure how I would go about calculating his velocity at that point. Energy conservation didn't seem to work, but then again I may still be missing something simple in that case. Thanks.

The bull does not have to be a mass on a spring to move like a mass on a spring. You are told the bull moves in simple harmonic motion with a cetain amplitude and frequency. For any harmonic motion, there is relationship between displacement, velocity, acceleration, amplitude and frequency that you will need for this problem. Where in an upward phase of motion does the bull start to slow down? How fast is it going at that point? How fast are you going? What is the bull's acceleration? What is your acceleration? What has to be true about the two accelerations for you to stay in contact with the bull?

You need to prove the statement I highlighted, or come to another conclusion by trying to do so.

 

[factor]

Thanks OlderDan. I realized what an idiot I was being...

-W^2Asin(Wt) = -9.81 is the point in time where the rider loses contact and of course the velocity at this point is given by the integral of the function at time t which of course is WAcos(Wt).