Oscillation

2778 days ago by MATH4R2013

m=mass of object (kg)
l=unstretched length of bungee cord (m)
h=height above ground
x=max elongation (from equilibrium)
   y=total elongation (elongation from the top)
g=gravitational acceleration (m/s/s) **assume positive and constant
f=time for free fall (s)
U=initial potential energy (Joules)
E=elastic potential energy (Joules)

Find the time it takes to undergo the free fall
Δl=Vit+.5gf^2
Since Vi=0
l=.5gf^2
f=(2l/g)^.5

l=11 g=9.8 f=(2*l/g)**.5 f 
       

Find the spring constant (k) of the bungee cord
KEi+PEi=KEf+PEf

We could use kinematics to find the kinetic energy but the initial potential energy is the same as the potential spring energy because the kinetic energies at those points are equal to zero and there is no outside work besides gravity being done on the system (assuming air resistance and friction are negligible)

0+U=0+E
U=m*g*h
E=.5*k*x^2

y=total elongation (elongation from the top)=height above ground-unstretched length

h=36 l=11 m=65 y=h-l k=(2*m*g*h)/(y^2) k 
       

Find the equilibrium point (o) of the subsequent oscillation
x=elongation (from the equilibrium point
Use Hooke's law
Fs=kx
Fg=mg

At any point the equilibrium point there is no net acceleration because the force of gravity is equal to the restoring force of the spring (**Free body diagram)

Fs=Fg ==> kx=mg

x=(m*g/k) x 
       
 
       

However, this means that the center of oscillation is (**8.68) meters below the highest point during the oscillation
Since the highest point of oscillation is (**11) meters below the building, the equilibrium position is (**19.68) meters below the building

Find the angular frequency (w) of the oscillation
w=(k/m)^.5

w=(k/m)^.5 w 
       

How long will it take to hit the ground?
t=f+(T/2)


Assume t=total time and T=period
w=2*pi*frequency
frequency=1/T

T=2*pi.n()*(m/k)^.5 T 
       
f=1.49829835452879 T=5.91344539630720 t=f+(T/4) print "The total time of fall is "+str(t)+"s" 
       
The total time of fall is 2.97665970360559s
The total time of fall is 2.97665970360559s

If the bungee jumper spots a thief, 50m away in the horizontal direction, running below at a constant velocity of 5 m/s, how long after the thief is spotted, should the jumper jump in order to knock the thief out?

WaitingTime=thiefTime-fallTime

d=50 v=5 wT=(d/v)-2.97665970360559 print "The jumper should wait "+str(wT)+"s" 
       
The jumper should wait 7.02334029639441s
The jumper should wait 7.02334029639441s