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3.7 Mechanical Vibrations

mass-spring-damper problem

Schematic of a mass suspended by a spring and a damper in parallel.

for now, let's leave the damper out

Schematic of a mass suspended only by a spring.

motion of mass traces out a sine or cosine

why?

Hooke's Law (spring force) :

\[ |F_s| = kL \]
  • k: spring constant
  • L: elongation with respect to the natural length of the spring

natural length: length of spring undisturbed → no force

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A time-lapse diagram showing a mass on a spring tracing a sine wave as it moves right.
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Two spring diagrams: one at natural length l, and one with a block attached causing elongation L.
  • \(l\): natural length
  • \(L\): elongation
    • \(L > 0 \rightarrow\) stretched
    • \(L < 0 \rightarrow\) compressed

direction of force of the spring is dependent on \(L\)

forces on the block:

Free body diagram showing upward force vector labeled spring |Fs|=KL and downward vector labeled weight=mg.

at equilibrium (leave the block hanging w/o stretching or compressing further)

\(\rightarrow\) forces balance out

\(KL = mg\)
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if we stretch it further

Diagram showing a spring-block system at equilibrium and then stretched further by a displacement u.

\(u = 0\) with respect to (length at equilibrium)

forces:

Free body diagram with an upward arrow labeled K(L+u) and a downward arrow labeled mg.

define \(u\) as positive down

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\[ |F_s| = k(L + u) \quad \text{if } u > 0 \rightarrow F_s < 0 \text{ (spring pulls up)} \]\[ u < L \text{ (compress)} \rightarrow F_s > 0 \text{ (spring pushes down)} \]

To account for this, we can just use:

\[ F_s = -k(L + u) \]

Newton's 2nd Law to describe the motion

force = mass · acceleration

\[ mu'' = \underbrace{mg}_{\text{weight}} - \underbrace{k(L + u)}_{\text{spring}} \]
\[ mu'' = mg - kL - ku \]

From earlier, we found:

\[ mg = kL \]
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Equation of Motion

So the equation of motion is:

\[ mu'' + ku = 0 \]

down is positive

Initial Conditions

  • \( u(0) = u_0 \) (initial elongation)
  • \( u'(0) = v_0 \) (initial velocity)

Example

Mass weighs 10 lb stretches a spring by 6 in. If the mass is pulled down a distance of 2 in from the equilibrium and set in motion with an upward velocity of 1 ft/s. Describe the motion \( u(t) \).

1st sentence: equilibrium condition

  • \( L = 6 \text{ in} = \frac{1}{2} \text{ ft} \)
  • Hooke's Law: \( |F_s| = kL \)
  • \( 10 \text{ lb} = k \cdot \frac{1}{2} \) (weight)
\[ k = 20 \text{ lb/ft} \]
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Simple Harmonic Motion: Mass-Spring System

\[mu'' + ku = 0\]

Note on Mass:

mass!

mass \(\neq\) weight

\(\text{weight} = m \cdot g = 10 \text{ lb}\)

\(g\) in English units is \(32 \text{ ft/s}^2\)

(\(\text{in SI units is } 9.8 \text{ m/s}^2\))

\[m = \frac{10}{32} = \frac{5}{16}\]
\[\frac{5}{16} u'' + 20 u = 0\]

Simplified Equation

\[u'' + 64 u = 0\]

Initial Conditions

\[u(0) = \frac{1}{6}\]
\[u'(0) = -1\]

“mass is pulled down 2 in”

\(2 \text{ in} = \frac{1}{6} \text{ ft}\)

negative because it's upward (down is positive)

Characteristic Equation & General Solution

\[r^2 + 64 = 0 \implies r = \pm 8i\]

\[u(t) = C_1 \cos(8t) + C_2 \sin(8t)\]

with initial conditions we can find \(C_1, C_2\)

Particular Solution

\[u(t) = \frac{1}{6} \cos(8t) - \frac{1}{8} \sin(8t)\]
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but it's much easier to analyze it in this form:

\[u(t) = R \cos(\omega_0 t - \delta)\]

\(R\)

Amplitude

(how much up/down)

\(\omega_0\)

angular frequency

(how fast)

\(\delta\)

phase shift

Conversion Process

how to convert \(u(t) = A \cos(\omega_0 t) + B \sin(\omega_0 t)\) to \(R \cos(\omega_0 t - \delta)\)?

identity:

\[\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)\]

Derivation

\(R \cos(\omega_0 t - \delta)\)

\(= R \cos(\omega_0 t) \cos(-\delta) - R \sin(\omega_0 t) \sin(-\delta)\)

(Note: \(\cos(-\delta) = \cos(\delta)\) and \(\sin(-\delta) = -\sin(\delta)\))

\(= R \cos(\delta) \cos(\omega_0 t) + R \sin(\delta) \sin(\omega_0 t)\)