Solutions to Applications of Higher Order Differential Equations (2)- Part A

1) A 5 lb. weight hangs vertically on a spring whose spring constant k = 10. The weight is pulled 6 inches farther down and released. Find the equation of motion, its period and the frequency. What would the units be for k?

Solution:

We have to keep in mind that British units are being used here so the acceleration of gravity, g = 32 ft/sec² and the mass m is in slugs. The mass, recall is W/ g so that:

 

Mass m = 5 lbs/ 32 ft/sec²     =    5/32 slug

 

 

We’re given that k = 10 so the initial equation of motion is written:

 

 

mx”  + 10x = mg

 

Now, divide through by the mass m (remember its value):

x”  + 10 x/ (5/32 sl) = 32

or:   x” + 64 x = 32

Recall in this harmonic motion eqn. the angular frequency is derived from:

w²  =     64   so that w =  Ö64  =  8

Or: w =  2pf  = 8 and f =  8/ 2p  =4/p    rad/sec

The period is just: T =    2p/w      =   2p/ 8 =  p/ 4  sec

 

Getting back to the solution: Given the above we may write:

x = c1 sin 8t + c2 cos 8t + ½

(Note that the 6” has been converted to feet which marks the added extension)

Now, before the extra 6” pull down what is the natural extension? Well, since the spring constant k = 10 i.e. 10lbs stretches the spring 1 ft. then 5 lbs. must stretch it 6” or ½’.

Therefore, at t = 0 at the instant of pull down we have a total extension of 1 ft so:

 

 

1 ft = c1 sin 8(0) + c2 cos (0) + ½

 

 

Or: 1 ft = 0 + c2 + ½

 

So: c2 = ½

 

Then the solution is:  x = ½ cos 8t + ½

What are the units of k?  From the preceding information, k is in lb/ft.