Download A mass of 25g is attached to a vertical spring with a

A mass of 25g is attached to a vertical spring with a spring constant k = 3
dyne/cm. The surrounding medium has a damping constant of 10 dyne*sec/cm. The mass is
pushed 5 cm above its equilibrium position and released.
Find (a) the position function of the
mass, (b) the period of the vibration, and (c) the frequency of the vibration.

x = position of the block

v = x' = velocity of the block

m = mass of the block = 25 g = 0.025 kg

k = spring constant = 3 dyne / cm = 3 ×10−5 N / 10-2 m = 3 ×10−3 N / m

B = damping constant = 10 dyne ·s / cm = 10×10−5 N· s/10-2 m = 10−2 N· s / m
Since we adjust the coordinate system so that x = 0 corresponds to the spring being
unstretched, then the stretch of the spring is simply equal to x. The spring force becomes
Fspring = − k x
In addition, there is a damping (friction) force that resists the motion. It is proportional to
the velocity. So we add Fdamping = −B v to get the total force
F = Fspring + Fdamping = − k x − B v
Combining this with Newton's law of motion F = m a, and the definition of acceleration
as the second derivative of position a = x'' we have the differential equation:
m x'' = −k x − b v or equivalently:
To simply the equation, we define the following parameters:
0 
k
3 103 N / m
3


rad / s
m
0.025kg
5
and

B 102 N  s / m

 0.2rad / s
2m 2  0.025kg
The differential equation now becomes: x  2 x  02 x  0
Continuing, we can solve the equation by assuming x  Ae t
Substituting this assumed solution back into the differential equation, we obtain:
 2  2  02  0
Solving for  , we find:      2  02
Since  2  02  0 ,  is complex, the system is under-damped.
     2  02    i 02   2
Substituting  back to the assumed solution x  Ae t
x  Ae
   2 02 t
 Ae t e
 i 02  2 t
 Ae t cos( 02   2 t )  iAe t sin( 02   2 t )
If we only take the real part,
x(t )  Ae  t cos(d t ) , where d  02   2  (
3 2
)  (0.2) 2  0.2 2  0.283rad / s
5
At the initial point t =0, the mass is 5 cm above its equilibrium.
x(0)  A  0.05m
So
(a) x(t )  0.05e0.2t cos(0.283t ) , the positive x direction is downward.
(b) The frequency of the vibration is f 
(c) The period of the vibration is T 
d 0.283rad / s

 0.045Hz
2
2
1
1

 22.214s
f 0.045Hz