# Oscillation Version 3

## 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)#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 show("f =") show(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) show("k = ") show(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) show("x = ") show(x) #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 show("w = ") show(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 show("T = ") show(T) f=1.49829835452879 T=5.91344539630720 t=f+(T/4) print "The total time of fall is "+str(t)+"s" #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? d=50 v=5 #WaitingTime=thiefTime-fallTime wT=(d/v)-2.97665970360559 print "The jumper should wait "+str(wT)+"s" #U=initial potential energy (Joules) #E=elastic potential energy (Joules)
 \newcommand{\Bold}{\mathbf{#1}}\verb|f|\phantom{\verb!x!}\verb|=| \newcommand{\Bold}{\mathbf{#1}}1.49829835452879 \newcommand{\Bold}{\mathbf{#1}}\verb|k|\phantom{\verb!x!}\verb|=| \newcommand{\Bold}{\mathbf{#1}}73.3824000000000 \newcommand{\Bold}{\mathbf{#1}}\verb|x|\phantom{\verb!x!}\verb|=| \newcommand{\Bold}{\mathbf{#1}}8.68055555555556 \newcommand{\Bold}{\mathbf{#1}}\verb|w|\phantom{\verb!x!}\verb|=| \newcommand{\Bold}{\mathbf{#1}}1.06252529381658 \newcommand{\Bold}{\mathbf{#1}}\verb|T|\phantom{\verb!x!}\verb|=| \newcommand{\Bold}{\mathbf{#1}}5.91344539630720 The total time of fall is 2.97665970360559s The jumper should wait 7.02334029639441s The total time of fall is 2.97665970360559s The jumper should wait 7.02334029639441s