4.9 Free Mechanical Vibrations
Spring-Mass Oscillator
When the spring is not stretched and the mass m is at rest, the system is at equilibrium.
Forces Acting in the System
When the mass m is displaced from equilibrium the spring exerts a force, πΉπ πππππ, against the displacement
πΉπ πππππ = βππ¦
where y is the displacement of the mass, and k is the spring constant or stiffness. The system also experiences friction, given by
πππππππππ = βππ π
π π
Where π β₯ 0 is the damping coefficient.
Any other external forces such as gravitational, electrical, or magnetic forces will be lumped together as the known
function πΉππ₯π‘(π‘).
Using Newtonβs Second Law, we get the second order differential equation
ππ¦β²β² = βππ¦ β ππ¦β² + πΉππ₯π‘(π‘)
or
ππ¦β²β² + ππ¦β² + ππ¦ = πΉππ₯π‘(π‘)
Undamped, Free Motion If we consider the simplest case, where there is no friction (π = 0) and no external forces (πΉππ₯π‘(π‘) = 0) acting on the
system then we get the equation
ππ2π¦
ππ‘2+ ππ¦ = 0
To put in standard form, we divide by m to get
π2π¦
ππ‘2+ π2π¦ = 0
where π2 =π
π. This equation describes undamped free motion or simple harmonic motion
The resulting auxiliary equation is
π2 + π2 = 0
Which has complex roots
π = Β±ππ
Thus the general solution is
π¦(π‘) = π1 cos ππ‘ + π2 sin ππ‘
Notes:
When initial conditions are used to find π1 and π2, the resulting particular solution is called the
equation of motion.
The period, measured in seconds, is then
The frequency, or cycles per second, is
(aka natural frequency)
The circular frequency, is
Examples
1. A 20 kg mass is attached to a spring.
a. If the frequency of simple harmonic motion is 2 πβ cycles/s, what is the spring constant k?
b. What is the frequency of simple harmonic motion if the original mass is replaced with an 80 kg
mass?
Alternate Form of π(π) A simpler form of π¦(π‘) that makes it easier to determine amplitude is
π¦(π‘) = π΄ sin(ππ‘ + π)
Where π΄ = βπ12 + π2
2 and π is the phase angle defined by tan π =π1
π2
Note: Be careful when solving for the phase angle. Remember that the domain of tanβ1 π₯ is restricted to the
first and fourth quadrants.
2. A 3-kg mass is attached to a spring with stiffness π = 48 π πβ . The mass is displaced 1 2β π to the left
of the equilibrium point and given a velocity of 2 m/sec to the right. The damping force is negligible.
Find the equation of motion of the mass along with the amplitude, period, and frequency. How long
after release does the mass pass through the equilibrium position?
3. Express the equation in the alternate form.
π¦(π‘) =2
3cos 2π‘ β
1
6sin 2π‘
Free damped motion
A mass will only experience free undamped motion in a perfect vacuum and is therefore not very realistic.
Thus on a damped system with no external forces we have
ππ2π¦
ππ‘2+ π
ππ¦
ππ‘+ ππ¦ = 0
The auxiliary equation associated with this equation is
ππ2 + ππ + π = 0
With roots
π =βπ Β± βπ2 β 4ππ
2π =
βπ
2πΒ±
βπ2 β 4ππ
2π
Underdamped or Oscillatory Motion
When the discriminant is negative we get two complex roots, πΌ Β± π½π, where
πΌ =βπ
2π πππ π½ =
β4ππ β π2
2π
And the general solution is
π¦(π‘) = ππΌπ‘(π1 cos π½π‘ + π2 sin π½π‘)
Or more simply
π¦(π‘) = π΄πΞ±t sin(π½π‘ + π)
Where π΄ = βπ12 + π2
2 and π is the phase angle defined by tan π =π1
π2
This system is called underdamped because there is not enough damping to prevent oscillation.
The coefficient π΄πβππ‘ is called the damped amplitude of vibration or damping factor.
Because this π¦(π‘) is not periodic, the quasi period, or time between maxima, is given by
And the quasi frequency is
Overdamped Motion
When the discriminant is positive we get two distinct real roots, π1 and π2, and a general solution
π¦(π‘) = π1ππ1π‘ + π2ππ2π‘
This system is called overdamped because the damping force is great enough to prevent oscillation.
Critically Damped Motion
When the discriminant is zero we get one real root, π1, and a general solution
π¦(π‘) = π1ππ1π‘ + π2π‘ππ1π‘
This system is said to be critically damped because any slight decrease in the damping force would result in oscillatory
motion. Critically damped motion is very similar to overdamped motion.
Example
4. A 1 kg mass is attached to a spring whose constant is 16 N/m, and the entire system is submerged in a
liquid that imparts a damping force numerically equal to 10 times the instantaneous velocity.
Determine the equation of motion if
a. The mass is initially released from rest from a point 1 meter to the right of the equilibrium
position.
b. The mass is initially released from a point 1 meter to the right of the equilibrium position with a
velocity of 12 m/s to left.