Physical Applications of 2nd ODE

Physical Applications of 2nd ODE
Simple Harmonic Motion
We consider the motion of an object with mass at the end of a spring that is either vertical or horizontal on a level surface as in a figure .

Hooke’s Law, says that if the spring is stretched units from its natural length, then it exerts a force that is proportional to x : restoring force =-kx
where k is a positive constant(called the spring constant). If we ignore any external resisting forces then, by Newton’s Second Law we have
m (d^2 x)/(dt^2 )=-kx or m (d^2 x)/(dt^2 )+kx=0
Example 1: A frictionless spring with a 10kg mass can be held stretched 1 meters beyond its natural length by a force of 40 N. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2.5 m/sec, find the position of the mass after t seconds.
Solution: From Hooke’s Law, the force required to stretch the spring is
40 =1k ? k=40
10x^ +40x=0 ; x(0)=1 and x^ (0)=2.5
10r^2+40=0 ? r=?2i
x=c_1 sin?2t+c_2 cos?2t
x(0)=1 ? c_2=1
x^ =2c_1 cos?2t-2c_2 sin?2t
x^ (0)=2.5 ? 2.5=2c_1 ? c_1=1.25
x=1.25 sin?2t+cos?2t

Damped Vibrations
We next consider the motion of a spring that is subject to a frictional force .An example is the damping force supplied by a shock absorber in a car or a bicycle. We assume that the damping force is proportional to the velocity of the mass and acts in the direction opposite to the motion. Thus:
damping force=-c dx/dt
where c is a positive constant, called the damping constant. Thus, in this case, Newton’s Second Law gives :
m (d^2 x)/(dt^2 )+c dx/dt+kx=0
Example 2: A spring with a mass of 2 kg has natural length 0.5 m. A force of 25.6 N is required to maintain it stretched to a length of 0.7 m. If the spring is immersed in a fluid with damping constant c=40 . Find the position of the mass at any time if it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.6 m/s.
Solution: 25.6 =0.2k ? k=128 ,m=2 and c=40
2 (d^2 x)/(dt^2 )+40 dx/dt+128x=0
2x^ +40x^ +128x=0 ; x(0)=0 and x^ (0)=0.6
x^ +20x^ +64x=0
r^2+20r+64=0 ?(r+4)(r+16)=0 ?r=-4 ,-16
x(t)=c_1 e^(-4t)+c_2 e^(-16t)
x(0)=0 ?c_1+c_2=0 ?(1)
x^ (t)=-4c_1 e^(-4t)-16c_2 e^(-16t)
-4c_1-16c_2=0.6 ?(2)
c_1=0.05 and c_2=-0.05
x(t)=0.05 ( e^(-4t)-e^(-16t) )

Electric Circuits
The circuit shown in Figure contains an electromotive force E, a resistor R , an inductor L, and a capacitor C, in series. If the charge on the capacitor at time t is Q=Q(t) , then the current is the rate of change of Q with respect to t : I=dQ?dt.

It is known from physics that the voltage drops across the resistor, inductor, and capacitor are
RI L dI/dt ( Q )/C
respectively. Kirchhoff’s voltage law says that the sum of these voltage drops is equal to
the supplied voltage
L dI/dt+RI+( Q )/C=E(t)
Since I=dQ?dt, this equation becomes
L (d^2 Q)/(dt^2 )+R dQ/dt+( Q )/C=E(t)
Example 3: A series circuit consists of a resistor with R=20 ?, an inductor with L=1 H, a capacitor with C=0.002 F, and a 12-V battery. If the initial charge and current are both 0, find the charge at time t.
. Solution:
(d^2 Q)/(dt^2 )+20 dQ/dt+( Q )/0.002=12
(d^2 Q)/(dt^2 )+20 dQ/dt+500 Q=12
r^2+20r+500=0
r=(-20??(400-2000))/2=-10?20i
Q_h=e^(-10t) (c_1 sin?20t+c_2 cos?20t )
Now Q_p=A ? dQ/dt=(d^2 Q)/(dt^2 )=0
So 500 Q_p=12 ? Q_p=( 3 )/125
Q(t)=e^(-10t) (c_1 sin?20t+c_2 cos?20t )+( 3 )/125
Q(0)=0 ? 0=c_2+( 3 )/125 ? c_2=-( 3 )/125
I=dQ/dt=e^(-10t) (?20c?_1 cos?20t-20c_2 sin?20t )-10e^(-10t) (c_1 sin?20t+c_2 cos?20t )
I(0)=0 ? 0=?20c?_1-10c_2 ?c_1=-( 3 )/250
Q(t)=e^(-10t) (-( 3 )/250 sin?20t-( 3 )/125 cos?20t )+( 3 )/125


1. A spring with a mass of 3 kg has damping constant c=30 and spring constant k=123. Find the position of the mass at time if it starts at the equilibrium position
with a velocity of 2 m/s.
2. A series circuit consists of a resistor with R=24 ?, an inductor with L=2 H,
a capacitor with C=0.005 F, and a 12-V battery. If Q(0)=0.001 and I(0)=0,
find the charge at time t.