Use the differential equation (W/g)y"
+ k2y' + k1y = F(t) where W represents
the weight of the object attached to the end of the spring, k2 is the
damping factor, k1 is the spring constant, F(t) is an external force
acting on the system, y gives the position of the bottom of the object as a function of time
with y = 0 the equilibrium position and y positive indicates below the
equilibrium position. Take g = 32.
To see a brief development of the governing equation given above and also
an applet go to the NTNU
Virtual Physics Laboratory, select "wave" from the Pop-up
menu and then select "Oscillation and wave".
A 64 pound weight is attached to the end of the
spring. After reaching the equilibrium position the spring is
stretched one foot below the equilibrium position. The weight is then released and
as it is released it is struck a downward blow giving it an initial velocity of
2 ft/sec. Take the moment the weight is released and struck as time
zero. At time zero a periodic external force given by F(t) = (1/2)cos(4t)
pounds begins acting on the system. t is time in seconds. Consider
the damping factor to be negligible, i.e., take k2 to be zero.
The spring constant is 32. Find the function giving y, the
position of the bottom of the weight as a function of time given in
seconds. Find y after four seconds. Find y after one hour.
Find y after one day based on the mathematical model. What is
going to happen to the spring? The graph below is an attempt by
computer software to show y as a function of time over the first 600 seconds. The
graph may be misleading since the function is periodic and would go
through almost 382 periods in 600 seconds. The graph does give a
sense of what is happening to the amplitude. Look at the second
graph below for more accuracy but a shorter time interval (about 20
periods). Use
it to check your answer. The graph on the right below shows one
fundamental period of the position function.
Find the function that gives the velocity of the object
attached to the end of the spring. What is wrong with the mathematical
model for large t? You may be helped looking at this
animation
of the action of the spring and weight as t goes from 0 to 80pi
seconds (Quicktime Version). In the animation the spring is 10 feet long in the
equilibrium position and the cross section of the weight is a circle with
a diameter of 2 feet. When t reaches 80pi in the animation, the
animation repeats itself starting again from t = 0. The picture at the right shows the position of
the spring and weight at time t = 0 seconds.
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Resonance
in Three Dimensions:
Tacoma
Narrows Bridge Collapse
Long Version
Wind Version
Animation Corrected by Bryan Paul
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