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

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Debye–Hückel equation wikipedia, lookup

Newton's law of universal gravitation wikipedia, lookup

Newton's laws of motion wikipedia, lookup

Equations of motion wikipedia, lookup

Partial differential equation wikipedia, lookup

Schrödinger equation wikipedia, lookup

Equation of state wikipedia, lookup

Van der Waals equation wikipedia, lookup

Exact solutions in general relativity wikipedia, lookup

Differential equation wikipedia, lookup

Schwarzschild geodesics wikipedia, lookup

Calculus of variations wikipedia, lookup

Two-body problem in general relativity wikipedia, lookup

Transcript
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