Child bouncing on pogo stick-harmonic motion

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The discussion focuses on estimating the stiffness of a pogo stick spring for a child weighing 35 kg who bounces every 2 seconds. Participants clarify the use of harmonic motion equations, emphasizing the importance of understanding angular speed and acceleration in simple harmonic motion. The correct approach involves using the relationship ω = √(k/m) to derive the stiffness constant k. There is confusion about the application of kinematics versus harmonic motion equations, with a consensus that the latter is more appropriate for this problem. Ultimately, the key takeaway is the need to apply the defining equation for simple harmonic motion, a = -ω²x, to solve for the spring's stiffness.
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Homework Statement


Estimate the stiffness of a spring in a childs pogo stickif the child has a mass of 35kg and bounces every 2 sec.


Homework Equations


x=Asin(wt)
w=2(pi)f
w=2pi/T
F=mg-kx=ma
U=1/2kx2
E=1/2mv2+1/2kx2


The Attempt at a Solution

The equations that are listed are the only-harmonic motion- equations that I am allowed to use. This problem seems much more complicated to me than it appears.Am i supposed to use kinematics equations first, then subtract the time the child is in the air from the two seconds? Also the child is going to be in contact with the ground for some specified period of time while the spring is contracting, and rebounding. If i just think about it as simple harmonic motion then the period is 2 seconds. Then w=2pi/2 or w=pi then i looked at the sin graph and noted that at pi/2 v=0 so if v=0 at pi/2 then v=Awcos(wt) t=1 ... so at pi/2 on the sin graph the graph is at max height so i figured that the pogo stick was in the air so acceleration=9.8m/s2 then i solved for A using the equation a=-Aw2sin(wt) and got A=.993m...am i doing this correct? Isnt there a better way of doing this?
 
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In simple harmonic motion, the acceleration is never constant. So you can never use the kinematics equation which only applies when the acceleration is constant.

The stiffness of the spring is k.

A is the amplitude of the oscillation and the way you are using the equation is totally wrong.

What you need to do is either remember what is the angular speed formula for springs or derive it from first principles.
 
i thought it was w=2(pi)f and w=2pi/T?
 
oh ok i just used F=-kx=ma so -k[Asin(wt)]=m[-Aw2sin(wt)]
so k=-345kg/s2
 
yes, \omega=\frac{2\pi}{T}, but there is a formula for the angular speed of a spring in shm.

For example, for a pendulum oscillating in small angles,

\omega=\sqrt{\frac{g}{l}}

you need to find a similar expression for omega for springs.
 
pb23me said:
oh ok i just used F=-kx=ma so -k[Asin(wt)]=m[-Aw2sin(wt)]
so k=-345kg/s2

Yes, correct. It would have been simpler if you realize that for springs,

\omega=\sqrt{\frac{k}{m}}
 
could i have derived that with the equations that i listed??
 
You need one more equation. The general equation for shm!
 
And jus what exactly is this equation that you speak of...
 
  • #10
a=-\omega^2x

This is the defining equation for shm.
 
  • #11
thanx
 
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