Harmonic motion of a bungee jumper

In summary, the bungee jumper undergoes simple harmonic motion with an amplitude of 6.9 meters and a frequency of 0.115 Hz. The equation for this motion is x=Acos(ωt). To find the velocity of the jumper at 0.25 seconds, the equation must be solved for x. The angular velocity is found by multiplying the frequency by 2pi and then plugging in the values.
  • #1
Capncanada
53
0

Homework Statement



A bungee jumper undergoes simple harmonic motion with amplitude 6.9 m and frequency 0.115 Hz. Assume the bungee jumper follows the simple harmonic motion equation x=A\cos (\omega t).

Find the velocity of the jumper at 0.25s

Homework Equations



x=Acos(\omega*t)

The Attempt at a Solution



Found the angular velocity by multiplying the frequency by 2pi,
then plugged my values into the harmonic motion equation to find the x value, and divide by the time interval to find the velocity, but the homework says that's the wrong answer.
 
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  • #2
Please explain the angular velocity aspect of this problem. Secondly, the question asks for the velocity at 0.25 seconds. They do not ask for average velocity.
 
  • #3
Would the angular velocity be the bungee sway as the jump goes on? I don't really understand it to be honest... Are you saying we don't use the given equation in this question?
 
  • #4
No, I was questioning your terminology. I do not think they are referring to the swaying motion of the bungee. In my opinion the solution supplied was for the up and down motion. But nevertheless, let's get on with the problem.

They supply you with a function for position. It has the variable time in it. They are looking for the velocity. So how does one get velocity from a position function?
 
  • #5
By dividing the position by time
 
  • #6
Capncanada said:
By dividing the position by time

No, that is not correct. If you divide the CHANGE in position by time you get AVERAGE velocity over that period of time. But the problem is not asking for that. The problem is asking for the velocity when the time is 0.25 seconds. The velocity is not constant in this problem.

So what do you do?
 
  • #7
I can't think of how to obtain velocity from the position function.
 
  • #8
Capncanada said:
I can't think of how to obtain velocity from the position function.

Have you had a course in calculus?
 
  • #9
I have, I just can't recall how to obtain the value needed here, which is why I'm coming for help.
 
  • #10
OK, is the function written correctly?

x=Acos(\omega*t)

What is the back slash? Typo?

If you have a relationship that gives the position as a function of time, the derivative of it gives the velocity. So take dx/dt.
 
  • #11
It's x=Acos(ωt), I should've clarified.

We've never had to take any derivatives for this class, but I'll try.
 
  • #12
Are you sure there's not a simpler way to solve this? I don't know how to derive this down...
 
  • #13
Capncanada said:
Are you sure there's not a simpler way to solve this? I don't know how to derive this down...

Not for the instantaneous velocity. The derivative of the cosine function is the negative sine function.

V = d (A*cos(omega*t))/dt = -omega * A * sin(omega * t)

So now all you have to do is plug in the values.
 
  • #14
Okay, so for the instantaneous velocity you need to derive the function with a position function. I got the correct answer, thank you.

My calculus teacher was really sub-par, I barely passed the class, had to do a ton of tutoring. That was a year ago, need to scrub up I guess.
 
  • #15
You're welcome. Calculus is very important in technical fields such as engineering, physics, etc. What are you studying?
 
  • #16
I'm a biology major, physics is a major requirement for me to graduate. I'm good with the other sciences, not so much in mathematics.
 
  • #17
Good luck with your studies and have a successful career. Signing off now.
 

1. What is harmonic motion?

Harmonic motion is a type of periodic motion in which a system or object repeats the same motion over and over again, with each cycle taking the same amount of time.

2. How does harmonic motion apply to bungee jumping?

When a bungee jumper jumps off a platform, they experience harmonic motion as they oscillate up and down due to the forces of gravity and the bungee cord. The motion of the jumper can be described by a sinusoidal curve.

3. What factors affect the harmonic motion of a bungee jumper?

The length of the bungee cord, the weight of the jumper, and the gravitational force are all factors that can affect the harmonic motion of a bungee jumper. The stiffer the bungee cord, the faster the jumper will oscillate, and the lighter the jumper, the higher they will bounce.

4. How is the frequency of the harmonic motion calculated for a bungee jumper?

The frequency of the harmonic motion can be calculated using the formula f = 1/T, where f is the frequency and T is the period, or the time it takes for one complete oscillation. The period can be measured by timing how long it takes for the bungee jumper to complete a certain number of oscillations.

5. Are there any safety concerns related to the harmonic motion of a bungee jumper?

Yes, there are safety concerns related to the harmonic motion of a bungee jumper. If the bungee cord is too long or too short, it can cause the jumper to hit the ground or rebound too high, potentially causing injury. It is important for bungee jumping operators to properly calculate and set the length of the bungee cord for each jumper based on their weight and the height of the jump.

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