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Mechanical Vibrations

The motion of a mass attached to a spring serves as a relatively simple example of the vibrations that occur in more complex mechanical systems. For many such systems, the analysis of these vibrations is a problem in the solution of linear differential equations with constant coefficients. Consider a body of mass m attached to one end of an ordinary spring that resists compression as well as stretching; the other end of the spring is attached to a fixed wall. According to Hooke’s Law, the restorative force FS that the spring exerts on mass is proportional to the distance x that the spring has been stretched or compressed. Because this is the same as the displacement x of the mass m from its equilibrium position, it follows that FS = −kx k > 0 is the spring constant. FS < 0 when x > 0 and FS > 0 when x < 0. The figure shows a mass attached to a dashpot – a device, like a shock absorber, that provides a force directed opposite to the instantaneous direction of the mass m. Assume that the dashpot is so designed that this force FR is proportional to the velocity dxdt

of the mass.

FR = −cv = −cdxdt

c > 0 is the damping constant of the dashpot. In general, FR is the frictional forces such as air resistance. In addition, the mass is subjected to given external forces FE = F t( ) . Then the total force acting on the mass (governing the motion of mass) is F = FS +FR +FE

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Using Newton’s Law,

F =ma =m d 2x

dt2=m !!x

m !!x + c !x + kx = F t( )

• If there is no dashpot – we ignore all frictional forces

c = 0 Motion is undamped.

• If c > 0 , damped motion. If there are no external forces, then the motion is free. The motion is forced if F t( ) ≠ 0 . Thus, m !!x + c !x + kx = 0 describes the free motion of a mass on a spring with dashpot but no external forces applied. The weight of the mass w =mg of the mass would stretch the spring a distance s0 determined by FS = −kx with FS = −w and x = s0 . Then

mg = ks0

s0 =mgk

This gives the static equilibrium position of the mass. If y denotes the displacement of the mass in motion, measured downward from its static equilibrium position, then m !!y + c !y + ky = F t( )

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Free Undamped Motion

m !!x + kx = 0

!!x +kmx = 0

R2 + km= 0

R2 = − km

R = ± −km

x t( ) = Acos ω0t( )+Bsin ω0t( )

C = A2 +B2 cosα = AC

sinα = BC

x t( ) =C ACcos ω0t( )+ B

Csin ω0t( )

#

$%

&

'(

=C cosα cos ω0t( )+ sinα sin ω0t( )( )=C cos ω0t −α( )

where C is the amplitude, ω0 is the circular frequency (in radians per second) and α is the phase angle.

T = 2πω0

where T is the period (the time needed to complete one complete cycle) and the frequency (the number of complete cycles per second) is

1T=ω0

2π

We have just derived the formula for simple harmonic motion.

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Free Damped Motion

m !!x + c !x + kx = 0

!!x +cm

!x + km= 0

ω0 =km

p = c2m

!!x + 2p !x +ω02x = 0

R2 + 2pR+ω02 = 0

R =−2p± 4p2 − 4ω0

2

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The solutions to this depend upon the discriminant.

p2 −ω02 =

c2

4m2 −km=c2 − 4km4m2

Overdamped case

c2 − 4km4m2 > 0

c2 − 4km > 0c2 > 4kmc2

4km>1

Therefore, the equation has two distinct real zeroes r1 and r2, both negative which gives you the solution x t( ) = c1er1t + c2er2t Since r1 and r2 are both negative, t→∞ x t( )→ 0

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Critically damped case

c2 − 4km = 0c2 = 4kmc2

4km=1

Therefore, the equation has one real root with a multiplicity of two which gives you the solution x t( ) = c1e− pt + c2te− pt = e− pt c1 + c2t( ) t→∞ x t( )→ 0 Underdamped case

c2 − 4km < 0c2 < 4kmc2

4km<1

Therefore the equation has 2 complex non-real zeroes.

R = −p± i ω 02 − p2

x t( ) = e− pt c1 cos ω1t( )+ c2 sin ω1t( )( )

ω1 = ω02 − p2 = 4km− c2

2mx t( ) =Ce− pt cos ω1t −α( )

C = c12 + c2

2

tanα = c2c1

This represents the exponentially damped oscillations of the body about its equilibrium point. The graph of this function lies between the graphs of y =Ce− pt and y = −Ce− pt . The graph touches of the solution touches these graphs when ω1t −α is a multiple of π . The

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motion is not periodic but it is useful to call ω1 the circular frequency with T1 =2πω1

the

pseudo-period of oscillation. Ce− pt is the time varying amplitude. The action of the dashpot has at least 3 effects:

1. It exponentially damps the oscillation in accord with the time-varying amplitude. 2. It slows the motion, that is, the dashpot decreases the frequency of motion. 3. It delays the motion.

m !!x + c !x + kx = F t( ) Machines with rotating components commonly involve mass-spring systems or their equivalents in which the external force is simple harmonic. F t( ) = F0 cos ωt( ) or F t( ) = F0 sin ωt( ) where F0 is the amplitude of the periodic force and ω is its circular frequency. Consider a car with rotating vertical flywheel. The cart has mass m−m0 , not including the flywheel of mass m0 . The centroid of the flywheel is off-center at distance α from its center and its angular speed is ω radians per second. The cart is attached to a spring (with constant k). Assume that the centroid of the cart itself is directly beneath the center of the flywheel and denote x(t) its displacement from its equilibrium position (where the spring is unstretched). The displacement x of the centroid of the combined cart and flywheel is given by

x =

m−m0( ) x +m0 x +α cos ωt( )( )m

= x + m0αmcos ωt( )

Let’s ignore friction and apply Newton’s Second Law m !!x = −kx

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because the external force exerted by the spring is −kx .

m !!x −m0αω

2 cos ωt( ) = −kxm !!x + kx =m0αω

2 cos ωt( )

Such a system is a reasonable model of a front-loading washing machine with clothes being washed loaded off-center. Undamped Foreced Oscillations

m !!x + kx = F0 cos ωt( )

!!x +kmx =

F0 cos ωt( )m

ω0 =km

!!x +ω02x = 0

R2 +ω02 = 0

R = ±ω0ixc = c1 cos ω0t( )+ c2 sin ω0t( )

ω0 is the natural frequency of spring system Assume initially that

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ω ≠ω0

xp = Acos ωt( )+Bsin ωt( )

xp" = −Aω sin ωt( )+Bω cos ωt( )

xp"" = −Aω2 cos ωt( )−Bω 2 sin ωt( )

−m ω 2Acos ωt( )+Bω 2 sin ωt( )( )+ k Acos ωt( )+Bsin ωt( )( ) = F0 cos ωt( )B = 0−mω 2Acos ωt( )+ kAcos ωt( ) = F0 cos ωt( )

A = F0k −mω 2 =

F0m

ω02 −ω 2

xp =

F0m

ω02 −ω 2 cos ωt( )

x = xc + xp

x t( ) = c1 cos ωt( )+ c2 sin ωt( )+F0m

ω02 −ω 2 cos ωt( )

In practice, a mechanical system with very little damping can be destroyed by resonance vibrations. An example can occur when a column of soldiers marches in step over a bridge. Any complicated structure such as a bridge has many natural frequencies of vibration. If the frequency of the soldiers cadence is approximately equal to one of the natural frequencies of the structure, the resonance will occur. Indeed the resulting resonance vibrations can be of such large amplitude that the bridge will collapse. Broughton Bridge near Manchester, England in 1831 is an example of this and it is the reason for the now standard practice of breading cadence when crossing a bridge. (http://www.information-britain.co.uk/famdates.php?id=923) Resonance may have been involved in the 1981 Kansas City disaster in which a hotel balcony (skywalk) collapsed with dancers on it. (http://skywalk.kansascity.com/articles/sections/disaster/)