Solutions to Differential Equation Application Problems

 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.

2) For the circuit shown, consider the special case of NO capacitor. Then the differential equation of interest becomes:


L (dI/dt) + RI = E


It’s instructive to first solve this form where E is presumed constant. Begin by re-arranging the DE to get:

L dI = (E – RI) dt so that:  ò LdI/ (E – RI) = ò  dt

Let u = (E – RI) ,    Then: du = - RdI and dI = -du/ R

So:   - L/R  ò du/u = ò  dt

 

 

Now, u = E – RI and this has a (+ve) value (e.g. uo = E) at t = 0 and we assume it remains positive so çu ç = u and:

 

 

(-L/R) ln u  = t + C

 

 

Or: -(L/R) ln (E – RI)  = t + C

 

 

The constant of integration C can be found knowing that I = 0 at t = 0 hence:

 

 

C = - (L/R)  ln E

 

After suitable substitutions and some algebra:

 

L/R [ln E – ln (E – RI)] = t

 

ln [E/ E – RI]  = Rt/ L

 

Taking natural logs:

 

E/ (E – RI)  = exp (Rt/L)

 

 

Which can also be written:  (E – RI)/ E = exp (- Rt/L)

Solving for I:   I = E/R [1 - exp (-Rt/L)]

 

 

Then the current (for t > 0) always increases toward the steady state value of E/R. Now, what about the solution if we let E = E_o sin wt? (Oscillatory voltage). 

 

 

It should be fairly clear from past articles –blog posts that the solution must have the form: A sin wt + B cos wt

 

Using the integrating factor, the solution we obtain (note this is not a higher order DE!) is:

 

I exp (at)  =  ò exp (at)  E dt  + C

 

Then one will obtain:

I =   E_o { RL sin wt -  wL ² cos wt}/ (R² +  w² L ²   )  +  k exp (-Rt/L)

 

3)   A ball weighing 0.75  lb. is thrown vertically upward from a point 6 ft above the surface of the Earth, with an initial velocity of 20 ft/ sec. As it rises it is acted upon by air resistance which is numerically equal to v /64 in pounds.  Where v is the velocity in ft. per second.  How high will the ball rise? 

Solution:

The net force is:  - v/ 64  - 3/4  =  m dv/dt

The ball has weight:  w = m/g  =   m/ (32 f/s/s)  =  (3/4)/  32   

So we need to solve the differential equation:

 (3/4)/ 32 (dv/dt)  =   - v/64   - 3/4   

And note initial value: v(0) = 20 f/s  

So the DE is simplified to: 

dv/dt + 2v/ 3  = -32   

The integrating factor is:  e ^2t/3   

So:  d / dt (e ^2t/3    v)   -32 e ^2t/3 

e ^2t/3    v   =   -32 (3/2)  e ^2t/3    +  C

v(t) =   = 48  +    C e ^-2t/3

To find C:

v(0) =  20    =   - 48  +  C  -  20

C  =   68,    Then:

v(t) =   = 48  +  68  e ^-2t/3

Setting v(t) = 0 gives the time at which the velocity is 0, which is the peak altitude,

48 =   68  e ^{- 2t/3}

But:   ln (48/68)  =   -2t/  3

Or:   t  =   - 3/2 ln (48/68)   = 0.522 s  

Now, to get the height we solve: 

ds/ dt = v = -48 + 68 e ^-2t/3

Integrating:  s(t) =  -48t -   68 (3/2)e ^-2t/3   +   C1   

We know at t = 0, s = 6 ft  or s(0)=  6

Then:  s(0)  = 68 (3/2)e ^-2t/3   + C1

 s(0)  = - 102    + C1   Or:   C1  =  108  

Finally:

s(t) =  -48t -   102  e ^-2t/3   +   C1   

For t = 0.522 s

s(0.522) =  -48 (0.522)  - 102 [exp (-0.348)]   +  108

s(t) =   10.922  ft.