3. Figure 1 shows the position vs. time curve for a backpack hanging off of a tr

Question

3. Figure 1 shows the position vs. time curve for a backpack hanging off of a tree by a bungee cord. The backpack has a mass of 30 kg. (a) What are the period and frequency of the system? (b) What is the stiffness (spring constant) of the bungee cord? (c) What is the maximum speed of the backpack? Roughly when does it occur? (d) When the backpack is 2.00 cm below its equilibrium position how fast is it going? (e) With the methods we have learned for writing down the position as a function of time for an oscillator, why are you unable to write down the position as a function of time? (f) Make the necessary change (you might need to redefine something) and then write down the position as a function of time for the backpack. (g) If we take the backpack and bungee cord to the moon (an expensive experiment...) and hang it on a tree there (finding a tree might be difficult...) will the frequency of oscillation change? Explain.

Explanation / Answer

a) The period of the system is the difference between t=1s and t=4s, so the period (T) is 3 s. The frequency is 1/T, so it is 1/3 s -1.(graphic values)

b) The spring constant (k) is equal to: k=w2*m....So first we have to find w (angular frequency)

w=2*pi*f=2*pi*(1/3)=2.094 rad/s

k=2.0942*30=131.595 Kg/s2=131.595 N/m

c) Vmax=A*w=0.06m*2.094 rad/s=0.126 m/s. It occurs when the mass is at the equilibrium position, we can roughly say that it occurs in 1 s (because we can see that the equilibrium position happens when the curve reaches y=0 and the total compresion or extension which is y=0.06 happends 1 s later)

d)V= +/- w* sqrt(A2- x2)

Where x is the elongation...so if it is 2 cm below its equilibrium position, it has a 2 cm elongation

V=2.094 rad/s *sqrt((0.06m)2-(0.02m)2)=0.118 m/s

e)Because we dont have the initial phase (IP)

f) x(0)=A*cos(wt+IP)....X(0)=0.06*cos(IP)....IP=acos(x/0.06).....Because we don't have the initial position for t=0 lets say that the intitial time is going to be (t=-2), so X=0.06 and IP=0

X=0.06*cos(2.094t)

g)No it wont change because T does not depends of gravity.

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