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, 𝐹𝑠𝑝𝑟𝑖𝑛𝑔, against the displacement

𝐹𝑠𝑝𝑟𝑖𝑛𝑔 = −𝑘𝑦

where y is the displacement of the mass, and k is the spring constant or stiffness. The system also experiences friction, given by

𝑭𝒇𝒓𝒊𝒄𝒕𝒊𝒐𝒏 = −𝒃𝒅𝒚

𝒅𝒕

Where 𝒃 ≥ 0 is the damping coefficient.

Any other external forces such as gravitational, electrical, or magnetic forces will be lumped together as the known

function 𝐹𝑒𝑥𝑡(𝑡).

Using Newton’s Second Law, we get the second order differential equation

𝑚𝑦′′ = −𝑘𝑦 − 𝑏𝑦′ + 𝐹𝑒𝑥𝑡(𝑡)

or

𝑚𝑦′′ + 𝑏𝑦′ + 𝑘𝑦 = 𝐹𝑒𝑥𝑡(𝑡)

Undamped, Free Motion If we consider the simplest case, where there is no friction (𝑏 = 0) and no external forces (𝐹𝑒𝑥𝑡(𝑡) = 0) acting on the

system then we get the equation

𝑚𝑑2𝑦

𝑑𝑡2+ 𝑘𝑦 = 0

To put in standard form, we divide by m to get

𝑑2𝑦

𝑑𝑡2+ 𝜔2𝑦 = 0

where 𝜔2 =𝑘

𝑚. This equation describes undamped free motion or simple harmonic motion

The resulting auxiliary equation is

𝑟2 + 𝜔2 = 0

Which has complex roots

𝑟 = ±𝜔𝑖

Thus the general solution is

𝑦(𝑡) = 𝑐1 cos 𝜔𝑡 + 𝑐2 sin 𝜔𝑡

Notes:

When initial conditions are used to find 𝑐1 and 𝑐2, 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 𝜋⁄ 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 𝒚(𝒕) A simpler form of 𝑦(𝑡) that makes it easier to determine amplitude is

𝑦(𝑡) = 𝐴 sin(𝜔𝑡 + 𝜑)

Where 𝐴 = √𝑐12 + 𝑐2

2 and 𝜑 is the phase angle defined by tan 𝜑 =𝑐1

𝑐2

Note: Be careful when solving for the phase angle. Remember that the domain of tan−1 𝑥 is restricted to the

first and fourth quadrants.

2. A 3-kg mass is attached to a spring with stiffness 𝑘 = 48 𝑁 𝑚⁄ . The mass is displaced 1 2⁄ 𝑚 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.

𝑦(𝑡) =2

3cos 2𝑡 −

1

6sin 2𝑡

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

𝑚𝑑2𝑦

𝑑𝑡2+ 𝑏

𝑑𝑦

𝑑𝑡+ 𝑘𝑦 = 0

The auxiliary equation associated with this equation is

𝑚𝑟2 + 𝑏𝑟 + 𝑘 = 0

With roots

𝑟 =−𝑏 ± √𝑏2 − 4𝑚𝑘

2𝑚 =

−𝑏

2𝑚±

√𝑏2 − 4𝑚𝑘

2𝑚

Underdamped or Oscillatory Motion

When the discriminant is negative we get two complex roots, 𝛼 ± 𝛽𝑖, where

𝛼 =−𝑏

2𝑚 𝑎𝑛𝑑 𝛽 =

√4𝑚𝑘 − 𝑏2

2𝑚

And the general solution is

𝑦(𝑡) = 𝑒𝛼𝑡(𝑐1 cos 𝛽𝑡 + 𝑐2 sin 𝛽𝑡)

Or more simply

𝑦(𝑡) = 𝐴𝑒αt sin(𝛽𝑡 + 𝜑)

Where 𝐴 = √𝑐12 + 𝑐2

2 and 𝜑 is the phase angle defined by tan 𝜑 =𝑐1

𝑐2

This system is called underdamped because there is not enough damping to prevent oscillation.

The coefficient 𝐴𝑒−𝜆𝑡 is called the damped amplitude of vibration or damping factor.

Because this 𝑦(𝑡) 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, 𝑟1 and 𝑟2, and a general solution

𝑦(𝑡) = 𝑐1𝑒𝑟1𝑡 + 𝑐2𝑒𝑟2𝑡

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, 𝑟1, and a general solution

𝑦(𝑡) = 𝑐1𝑒𝑟1𝑡 + 𝑐2𝑡𝑒𝑟1𝑡

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.