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5.4 Mechanical Vibrations

mass-spring-damper

Schematic of a mass m attached to a spring and damper, positioned on a frictionless ground with equilibrium at x=0.
  • block mass \( m \)
  • spring w/ spring constant \( k \)
  • damper (dashpot) w/ damping constant \( c \)
  • \( x = 0 \) is equilibrium
Ground (no friction)

Give \( x(0) \) and or \( x'(0) \), find \( x(t) \)

spring resists displacement: \( F_s = -kx \) (Hooke's Law)

\( x \): displacement from equilibrium

damper resists velocity: \( F_d = -cx' \)

Newton's 2nd Law: \( \sum F = mx'' \)

\[ -kx - cx' = mx'' \implies mx'' + cx' + kx = 0 \]

Linear constant coeff

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Example

  • mass \( 8 \text{ kg} \)
  • no damper
  • spring such that a force of \( 40 \text{ N} \) stretches it from equilibrium by \( 5 \text{ cm} \).
\[ mx'' + cx' + kx = 0 \]

\( m = 8 \)

\( c = 0 \)

\( k \) to be found

Hooke's Law: \( F_s = kx \)

\( x \): displacement from equilibrium

\( 40 \text{ N} = k \cdot (0.05 \text{ m}) \) (meters)

\( k = 800 \text{ N/m} \)

\[ 8x'' + 800x = 0 \] \[ x'' + 100x = 0 \]

\( r^2 + 100 = 0 \)

\( r = 10i, -10i \)

\[ x(t) = C_1 \cos(10t) + C_2 \sin(10t) \]

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Initial Conditions and Solution for Harmonic Motion

initial conditions: \( x(0) = 0 \), \( x'(0) = 10 \)

no initial displacement

initial velocity \( 10 \text{ m/s} \)

\[ x(t) = C_1 \cos(10t) + C_2 \sin(10t) \]

\[ x'(t) = -10 C_1 \sin(10t) + 10 C_2 \cos(10t) \]

\( x(0) = 0 \rightarrow 0 = C_1 \)

\( x'(0) = 10 \rightarrow 10 = 10 C_2 \rightarrow C_2 = 1 \)

\[ x(t) = \sin(10t) \]

A sine wave graph of x(t) versus t, starting at the origin with an amplitude of 1 and -1.
  • period:\( \frac{2\pi}{\text{freq}} = \frac{2\pi}{10} = \frac{\pi}{5} \text{ seconds} \)
  • frequency (circular):\( 10 \text{ rad/s} \)
  • amplitude:\( 1 \)
  • frequency (linear):\( \frac{1}{\text{period}} = \frac{5}{\pi} \text{ cycles per second (Hz)} \)
A simple mass-spring system diagram showing a block on a surface with a horizontal double-headed arrow.
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Harmonic Motion with Non-Zero Initial Displacement

Same setup, but \( x(0) = 2 \) and \( x'(0) = 10 \)

\[ x(t) = C_1 \cos(10t) + C_2 \sin(10t) \]

\[ x'(t) = -10 C_1 \sin(10t) + 10 C_2 \cos(10t) \]

\( x(0) = 2 \rightarrow \dots \rightarrow C_1 = 2 \)

\( x'(0) = 10 \rightarrow \dots \rightarrow C_2 = 1 \)

now particular solution is

\[ x(t) = 2 \cos(10t) + \sin(10t) \]

  • Same freq: \( 10 \text{ rad/s} \)
  • Same period: \( \frac{\pi}{5} \)
  • amplitude?

Alternate Form of the Solution

express in alternate form:

\[ x(t) = A \cos(\omega t) + B \sin(\omega t) \]
\( \rightarrow \)

\[ x(t) = C \cos(\omega t - \delta) \]

\( C \)

amplitude

\( \omega \)

omega (freq)

\( \delta \)

delta (phase shift)

using the identity \( \cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b) \)

after some algebra, we get

\[ C = \sqrt{A^2 + B^2} \]

\[ \delta = \tan^{-1}\left(\frac{B}{A}\right) \]

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here,

\[ C = \sqrt{2^2 + 1^2} = \sqrt{5} \]\[ \delta = \tan^{-1}\left(\frac{1}{2}\right) \approx 0.464 \]

\( x(t) = \sqrt{5} \cos(10t - 0.464) \)

amplitude \( \sqrt{5} \)

back to \( mx'' + cx' + kx = 0 \)

\( m, c, k \neq 0 \) w/ damper

\[ mr^2 + cr + k = 0 \]\[ r = \frac{-c \pm \sqrt{c^2 - 4km}}{2m} \]

\( c^2 - 4km \) is called the discriminant

it determines the type of solution

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if \( c^2 - 4km > 0 \) (\( c^2 > 4km \) or \( c > \sqrt{4km} \))

roots of char. eq are real and distinct

\[ x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]

strong damper \( \rightarrow \) overdamped

if \( c^2 - 4km = 0 \) (\( c^2 = 4km \))

roots are real and repeated

\[ x(t) = C_1 e^{rt} + C_2 t e^{rt} \]

damped "just right" \( \rightarrow \) critically damped

if \( c^2 - 4km < 0 \) (\( c^2 < 4km \))

roots : \( r = a \pm bi \)

\[ x(t) = C_1 e^{at} \cos(bt) + C_2 e^{at} \sin(bt) \]

weak damper \( \rightarrow \) underdamped

only case w/ oscillations

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Oscillatory Motion Analysis

The following graph illustrates the displacement function \( x(t) \) over time \( t \). The vertical axis represents the displacement, with key values marked at \( \sqrt{5} \), \( 2 \), \( 1 \), \( 0 \), \( -1 \), and \( -2 \). The horizontal axis represents time \( t \), with a major interval marked at \( 0.5 \) and \( 1 \).

A red sinusoidal wave graph of x(t) vs t, showing a peak near t=0.1 and t=0.7, with a period of approximately 0.6.

Phase Shift Observation

The initial peak of the wave is shifted from the origin. This horizontal displacement is noted as being affected by \( \delta \) (the phase constant).

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Comparison of Damping Regimes

This graph compares four different types of oscillatory behavior for a system \( x(t) \) over time \( t \). The vertical axis shows displacement from \( -20 \) to \( 20 \), and the horizontal axis shows time from \( 0 \) to \( 20 \).

Graph comparing undamped (black), underdamped (green), critically damped (blue), and overdamped (red) motion curves.

Damping Classifications

  • Undamped: Represented by the black curve, showing a continuous, constant-amplitude oscillation.
  • Underdamped: Represented by the green curve, showing decaying oscillations that cross the equilibrium axis.
  • Crit. damped (Critically Damped): Represented by the blue curve, returning to equilibrium as quickly as possible without oscillating.
  • Overdamped: Represented by the red curve, returning to equilibrium slowly without oscillating.
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Suspension Modeling

Suspension modeled as mass-spring-damper

Diagram of a car body of mass m with an upward force F, supported by a spring k and damper c above a ground surface.

The diagram illustrates a vehicle suspension system where the car body is represented as a mass \( m \). An external force \( F \) acts vertically upwards. The suspension components between the body and the ground are a spring with constant \( k \) and a damper with coefficient \( c \). The vertical displacement is denoted by \( x \).

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Mechanical and Electrical Analogies

Mass-Spring-Damper System

(a)

Mechanical diagram of a mass m connected to a damper c and spring k, with displacement y.
\[ my'' + cy' + ky = F(t) \]

\( F(t) \): External force

RLC Circuit

(b)

Electrical circuit diagram with a resistor R, capacitor C, inductor L, and voltage source E(t).
\[ LQ'' + RQ' + \frac{1}{C}Q = E(t) \]

\( Q \): charge

Note the mathematical similarity between the second-order differential equations governing mechanical and electrical systems.