Download 5.1 - Mass/Spring Systems

5.1.1 – Spring Mass Systems:
Free Undamped Motion Math 230
Warnock - Class Notes _______________ __________ states that a spring exerts a restoring force F opposite to the stretch of elongation s and ______________________ to that elongation. That is F  ks Where k is a constant of proportionality call the ___________ ____________. So if a mass weighed ________ pounds and stretches a spring _______ foot Then, After a mass is attached to a spring, it stretches the spring by an amount s to an ____________________ ____________ where it’s weight W is balanced by the restoring force F  ks . Weight is defined by ____________ times ______________. _______________ can be measured in and slugs _ g  32 ft/s2 kilograms _ g  9.8 m/s2 grams _ g  980 cm/s2 So we have mg  ks or mg  ks  0 If the mass is displaced by an amount x from its equilibrium position, the restoring force is k  x  s Newton’s 2nd Law says that Force = ____________ x __________________. Assuming there are no external forces on the system, called ________ _____________, we can equate Newton’s 2nd Law with the resultant force of the restoring force and the weight. So we have 2
m d 2x  mg  k  s  x   mg  ks  kx   kx dt
The negative sign here represents that the restoring force of the spring is acting opposite the direction of motion. We will represent the displacement ________________ the equilibrium position x  0 as _______________. Finally, dividing by mass, we have d 2x  k x  0 dt 2 m
or d 2x 2x  0 dt 2
where 2  k m
d 2x 2x  0 dt 2
where 2  k m
This equation is said to describe __________ ______________ ____________. Some clear initial conditions x  0  x0 ‐ initial displacement _ x  0  x1 ‐ initial velocity _ So, x0  0, x1  0 means 
When x 0  0 the mass is said to be _______________ _______ ________. 2x
d
To solve 2   2 x  0 , note that the solutions to its auxiliary dt
equation m2   2  0 are m1  i, m2  i Giving a general solution of x t   c1 cost  c2 sin t The ______________ of motion here is T
(Equation of Motion)  2 . This is the amount of time it takes to 
complete one full cycle (highest point back to highest point, or lowest point back to lowest point). The ______________ of motion is f  1   . This is the number of cycles T
2
completed each second. Amplitude is given by A 
c12  c22 (see page 195) Ex #1. Ex #2. A mass of 5kg, attached to a spring, stretches it 20 cm. Initially, the mass is released from a point 15 cm above the equilibrium position. Find the equation of motion. Ex #3. A mass weighing 64 lbs stretches a spring 0.32 feet. The mass is initially released from a point 8 in. above equilibrium with a downward velocity of 5 ft/s. Find the equation of motion.