Try using the method of separation of variables to solve the following:

dy/dx = y(cos(x)),      y(0) = 1

The graph of the solution is shown on the right.  Click on the graph to see an animation of the direction field vectors moving across the screen for increasing values of x along with an animated solution point.

Simple Harmonic Motion--The vibrating spring with no damping and no forcing function.

Find the position function for a 32 pound object attached to the end of a spring with a spring constant of 1 with no forcing function acting on the system and neglecting damping (k₂ = 0).  The object is pulled down until the spring is stretched to 5 feet below its equilibrium position and then the object is released from rest (i.e., the initial velocity is 0).  Click here to see an animation of the motion.  Take y = 0 to be the equilibrium position and y positive to indicate below the equilibrium position (contrary to the labeling in the animation).  Use the differential equation given below.

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 object as a function of time with y = 0 the equilibrium position and y positive indicates below the equilibrium position (contrary to the labeling in the picture on the right).  Take g = 32.  

A 64 pound weight is attached to the end of the spring.  After reaching the equilibrium position the spring is stretched one foot beyond 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 shows y as a function of time over the first 600 seconds.  It is misleading.  Why?

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.  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.

The vibrating spring with constant velocity over intervals

Find the governing differential equation and position functions for a 32 pound object attached to the end of a spring with a spring constant of 1 and a forcing function that yields a constant velocity in the direction of motion.  This velocity changes sign periodically.  The forcing function is piecewise.  The object is pulled down until the spring is stretched to 5 feet below its equilibrium position and then the object is released with an initial velocity of -1 ft/sec and the forcing function produces a constant velocity of -1 ft/sec.  After the object has traveled 10 feet it is impeded and reverses direction with an initial velocity at that point of 1 ft/sec. and the forcing function changes to produce a constant velocity of 1 ft/sec.  This behavior continues indefinitely.  Click here to see an animation of the motion with no scales and click here to see the animation with scales on the y-axis.  Take y = 0 to be the equilibrium position and y positive to indicate below the equilibrium position (contrary to the labeling in the animation).  Neglect damping.

A pendulum with a rod of length 32 feet is deformed pi/6 radians from the vertical position and released (initial velocity 0 at time t=0).  The governing differential equation is given below.  The picture at the right shows the initial position of the pendulum (just as it is released at time t=0).  Click on the picture to see an animation.  Click here to see an animation represented in a scaled coordinate system.  The animation corresponds to the solution to the linear model.

Here are two wonderful applets.  Simple pendulum is similar to my animation but allows you to interactively change the initial condition, follows the nonlinear model, and allows you to compare the fundamental periods for the linear and nonlinear models.  It is really quite nice.  Spring pendulum uses a spring for the rod.  There does not exist a differential equation able to model this motion.  Take a look.

PARACHUTE PROBLEM:  A man with a parachute jumps out of an airplane at an altitude of 5000 feet.  After 5 seconds his parachute opens and at the moment his parachute opens he catches a brief updraft.  The instant the first man's parachute opened a second man jumped out of the airplane at the same altitude.  The second man has the same weight and drag coefficient as the first man.  Regrettably the second man's parachute and back-up parachute both malfunctioned and did not open.

    animation1--The first man with no scales

    animation2--The first man with scales

    animation3--Both jumpers, no scales, with the animation starting the moment the second man jumps

    animation4--Both jumpers with scales

Mathematics can be used to prove that there will be a limiting velocity, i.e., that there will be a limit to how fast you will fall.  There will be a limit to how fast you will fall even if you are not wearing a parachute but the limit will be a much larger number. To see a description of this complete with an animation go to free fall compared to using a parachute.  Another example can be found in the US Naval Academy parachuting applet.  

Click here for a problem on parachuting including the development of the velocity function and a few hints and helpful graphs and a link to the solution.

In the animation the building is in green and the red circular path is that of the end of the pipe.  The blue point moving along the red circular path is indicating the speed the end point of the pipe would move at if it was moving at a constant speed and taking the required length of time to reach the top of the building.  The blue point on the small purple circle on top of the building is indicating the speed with which the winch is rotating to pull in the purple rope at a rate of -0.2m/sec.  The pipe is in blue.  Click here or on the picture at the right to see the animation.  Find the position function for the point at the end of the pipe.  

(a)  Write the velocity as a function of time if the wind is blowing at 20 ft/sec and after one second the boat is moving at 5 ft/sec.  Assume the boat started from rest.

(b)  Use the result in part (a) to write the distance traveled by the boat as a function of time.  

Click on the picture below to see an animation showing the first 8 seconds of the boat's travel starting from rest.  Click here to see the animation placed in a coordinate system.  Solution