Free and Forced Vibration

Dynamics · FE Reference Handbook section

Dynamics

20 formulas

10 exam-style examples

~60 min

All Dynamics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A single degree-of-freedom vibration system, containing a mass m, a spring k, a viscous damper c, and an external applied force
The equation of motion for the displacement of x is:
If the externally applied force is 0, this is a free vibration, and the motion of x is solved as the solution to a homogeneous
sum of the homogeneous solution and a particular solution.
For forced vibrations, one is typically interested in the steady state behavior (i.e. a long time after the system has started), which

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Acceleration from Newton's second law — Free and Forced Vibration

A 3168 lb cart is pulled horizontally by a 611 lb force against rolling resistance μ = 0.10. What is its acceleration?

Given

$W = 3168 lb$
$F = 611 lb$
$\mu = 0.10$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 3168/32.2 = 98.39 slugs$
Resistance — F_r = μW = 0.10(3168) = 316.8 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 611 − 316.8 = 294.2 lb
Acceleration — a = ΣF/m = 294.2/98.39 = 2.990 ft/s²

Answer:

a ≈ 2.99 ft/s²

Why the other options are there

0.093 ft/s² (weight used as mass)
6.21 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 2

Acceleration from Newton's second law — Free and Forced Vibration (2)

A 3332 lb cart is pulled horizontally by a 364 lb force against rolling resistance μ = 0.10. What is its acceleration?

Given

$W = 3332 lb$
$F = 364 lb$
$\mu = 0.10$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 3332/32.2 = 103.5 slugs$
Resistance — F_r = μW = 0.10(3332) = 333.2 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 364 − 333.2 = 30.8 lb
Acceleration — a = ΣF/m = 30.8/103.5 = 0.298 ft/s²

Answer:

a ≈ 0.30 ft/s²

Why the other options are there

0.009 ft/s² (weight used as mass)
3.52 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 3

Acceleration from Newton's second law — Free and Forced Vibration (3)

A 1873 lb cart is pulled horizontally by a 637 lb force against rolling resistance μ = 0.20. What is its acceleration?

Given

$W = 1873 lb$
$F = 637 lb$
$\mu = 0.20$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 1873/32.2 = 58.17 slugs$
Resistance — F_r = μW = 0.20(1873) = 374.6 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 637 − 374.6 = 262.4 lb
Acceleration — a = ΣF/m = 262.4/58.17 = 4.511 ft/s²

Answer:

a ≈ 4.51 ft/s²

Why the other options are there

0.140 ft/s² (weight used as mass)
10.95 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 4

Acceleration from Newton's second law — Free and Forced Vibration (4)

A 3801 lb cart is pulled horizontally by a 840 lb force against rolling resistance μ = 0.05. What is its acceleration?

Given

$W = 3801 lb$
$F = 840 lb$
$\mu = 0.05$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 3801/32.2 = 118.0 slugs$
Resistance — F_r = μW = 0.05(3801) = 190.1 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 840 − 190.1 = 650.0 lb
Acceleration — a = ΣF/m = 650.0/118.0 = 5.506 ft/s²

Answer:

a ≈ 5.51 ft/s²

Why the other options are there

0.171 ft/s² (weight used as mass)
7.12 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 5

Acceleration from Newton's second law — Free and Forced Vibration (5)

A 3005 lb cart is pulled horizontally by a 824 lb force against rolling resistance μ = 0.10. What is its acceleration?

Given

$W = 3005 lb$
$F = 824 lb$
$\mu = 0.10$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 3005/32.2 = 93.32 slugs$
Resistance — F_r = μW = 0.10(3005) = 300.5 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 824 − 300.5 = 523.5 lb
Acceleration — a = ΣF/m = 523.5/93.32 = 5.610 ft/s²

Answer:

a ≈ 5.61 ft/s²

Why the other options are there

0.174 ft/s² (weight used as mass)
8.83 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 6

Acceleration from Newton's second law — Free and Forced Vibration (6)

A 4900 lb cart is pulled horizontally by a 763 lb force against rolling resistance μ = 0.30. What is its acceleration?

Given

$W = 4900 lb$
$F = 763 lb$
$\mu = 0.30$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 4900/32.2 = 152.2 slugs$
Resistance — F_r = μW = 0.30(4900) = 1,470 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 763 − 1,470 = -707.0 lb
Acceleration — a = ΣF/m = -707.0/152.2 = -4.646 ft/s²

Answer:

a ≈ -4.65 ft/s²

Why the other options are there

-0.144 ft/s² (weight used as mass)
5.01 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 7

Acceleration from Newton's second law — Free and Forced Vibration (7)

A 1845 lb cart is pulled horizontally by a 415 lb force against rolling resistance μ = 0.15. What is its acceleration?

Given

$W = 1845 lb$
$F = 415 lb$
$\mu = 0.15$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 1845/32.2 = 57.30 slugs$
Resistance — F_r = μW = 0.15(1845) = 276.8 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 415 − 276.8 = 138.3 lb
Acceleration — a = ΣF/m = 138.3/57.30 = 2.413 ft/s²

Answer:

a ≈ 2.41 ft/s²

Why the other options are there

0.075 ft/s² (weight used as mass)
7.24 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 8

Acceleration from Newton's second law — Free and Forced Vibration (8)

A 2303 lb cart is pulled horizontally by a 990 lb force against rolling resistance μ = 0.25. What is its acceleration?

Given

$W = 2303 lb$
$F = 990 lb$
$\mu = 0.25$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 2303/32.2 = 71.52 slugs$
Resistance — F_r = μW = 0.25(2303) = 575.8 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 990 − 575.8 = 414.3 lb
Acceleration — a = ΣF/m = 414.3/71.52 = 5.792 ft/s²

Answer:

a ≈ 5.79 ft/s²

Why the other options are there

0.180 ft/s² (weight used as mass)
13.84 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 9

Acceleration from Newton's second law — Free and Forced Vibration (9)

A 2273 lb cart is pulled horizontally by a 779 lb force against rolling resistance μ = 0.25. What is its acceleration?

Given

$W = 2273 lb$
$F = 779 lb$
$\mu = 0.25$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 2273/32.2 = 70.59 slugs$
Resistance — F_r = μW = 0.25(2273) = 568.3 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 779 − 568.3 = 210.8 lb
Acceleration — a = ΣF/m = 210.8/70.59 = 2.986 ft/s²

Answer:

a ≈ 2.99 ft/s²

Why the other options are there

0.093 ft/s² (weight used as mass)
11.04 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration

Example 10

Acceleration from Newton's second law — Free and Forced Vibration (10)

A 3328 lb cart is pulled horizontally by a 310 lb force against rolling resistance μ = 0.20. What is its acceleration?

Given

$W = 3328 lb$
$F = 310 lb$
$\mu = 0.20$
$g = 32.2 ft/s^{2}$

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W/g = 3328/32.2 = 103.4 slugs$
Resistance — F_r = μW = 0.20(3328) = 665.6 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 310 − 665.6 = -355.6 lb
Acceleration — a = ΣF/m = -355.6/103.4 = -3.441 ft/s²

Answer:

a ≈ -3.44 ft/s²

Why the other options are there

-0.107 ft/s² (weight used as mass)
3.00 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Free and Forced Vibration