Unit III Lesson 7

One of these is ‘free motion’ of a spring or any vibrating object.

Free Undamped Motion (Oscillatory Motion)

Consider a mass m suspended on a spring from a fixed vertical support. Let L₀ be the natural length of the spring. When a weight is hung on the spring, it extends the spring an additional amount L, where it rests in equilibrium position. According to Hooke’s Law, the tension on the spring is directly proportional to this extension (F=kL) and this force would be the same as the weight of the object added m*g. Therefore kL=mg for m=the mass of the object and g=the gravitation constant. Now suppose at some fixed time t_0, the mass is displaced y₀ units from its equilibrium position. If y=y(t) is the distance of the mass from equilibrium at time t, then the extension at time t would be L+y(t). Note that downward direction is usually considered to be positive.

According to one of Newton’s laws of mechanics, the rate of change of downward momentum=the net downward force, in math symbols:

Noting the derivative of y’(t) is y’’(t) and that mg=kL this will simplify to:

You should already know how to produce the general solution:

If the system is set into motion at time t₀ by a displacement y₀ units and given an initial velocity v₀, we have the initial conditions y(t₀)=y₀ and y’(t₀)=v₀. The whole problem is now a second order linear IVP with constant coefficients.

This is an example of what is called free ‘undamped’ motion. This motion will go on with no change unless something occurs to alter it. An example in animated form, look at the following. (This animation was made by a Russian scientific group. Their site can be viewed at this link.)

Note that the graph showing its position appears to be a regular sine wave with a fixed amplitude. That amplitude is not obvious from our solution, but a little work with trig can produce it. If we start with a function of form Bcos(wx)+Csin(wx), note that this can be expressed A(B/A)cos(wx)+A(C/A)sin(wx). In particular, let A²=B²+C². Let z be an angle whose sine is B/A and whose cosine is C/A (i.e. tan(z)=B/C). Now we can write the function as

Asin(z)cos(wx)+Acos(z)sin(wx)=A(sin(z)cos(wx)+cos(z)sin(wx))

Which can be written (lookup the sin(A+B) identity):

Asin(wx+z)

Therefore our solution can be written as

where .

If you’d like to see this change, try comparing the graph of y=3cos(2x)+4sin(2x) with the graph of y=5sin(2x+tan^-1(3/4)) or better yet so you can see the difference compare y=3cos(2x)+4sin(2x) and y=5sin(2x+tan^-1(3/4))+1.

This graph was made with 'Derive' of the functions:

y=3cos(2x)+4sin(2x) and y=5sin(2x+tan^-1(3/4))+1. Note one is just 1 unit above the other.

Ex1. A 32 lb weight stretches a spring 2 feet. Determine the amplitude and period of motion if the weight is released 1 foot above equilibrium position with an initial upward velocity of 2ft/sec. How many complete vibrations will the weight have completed at the end of seconds?

Take this link after you’ve tried it yourself.

Damping and external force

Such a concept as free undamped motion is not realistic, although it appears in some applications. A more realistic approach would be to consider that there is a damping force acting on the weight that is proportional to its velocity at any given time. You see, there will be friction with the surrounding medium that slows it down, or maybe the system itself has a damping mechanical device. Think of the shock absorbers on your car. The springs don’t continually bounce up and down after you go over a bump since the ‘shocks’ act to dampen the bouncing. In either case the "Proportional to velocity" adds to the differential equation and it becomes:

for some positive constant B.

There could also be an external force f(t) acting on the weight. Perhaps the spring support is moving, or the weight could be attached to another moving system. In that case another force is added and the DE becomes the one we’ve been studying:

The solution to this equation will give the displacement y(t) of the weight from its equilibrium position at time t and the initial conditions are y(t₀)=y₀ representing the initial position of the weight and y’(t₀)=v₀ representing its initial velocity. If you want to experiment with this equation for various values of m, B and k, visit this website from the US military academy.

Ex2. A 4 foot spring measures 8 feet long after an 8 lb. weight is attached to it. The medium through which it moves offers a resistance numerically equal to times the instantaneous velocity. Find the equation of motion if the weight is released from the equilibrium position with a downward velocity of 5 ft/sec. Find the time at which the weight attains its extreme displacement from the equilibrium position.

Take this link after trying it yourself.

Electrical Circuits (LRC type)

If we denote the current in the circuit by i(t)and the charge on the capacitor as q(t) at time t, and L, R and C are inductance, resistance and capacitance, then according to Kirchoff’s Second Law, The impressed voltage E(t) on a closed loop must equal the sum of the voltage drops across the loop. Also since i(t) is related to charge on a capacitor by i(t)=dq/dt we have the second order linear DE:

Each of the voltage drops on the left are from the circuit and constants given:

Inductor =

Resistor =iR = R

Capacitor=

The solution q(t) will give the charge on the capacitor at time t.

An alternate form is to differentiate both sides of

and get the equation:.

Note h = henrys, =ohms,  f = farads, C = coulombs, A = amperes, V = volts.

Take this link after trying to solve it yourself.