4.9 Free Mechanical Vibrations

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Elaine Barnett
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1 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, F spring, against the displacement F spring = ky where y is the displacement of the mass, and k is the spring constant or stiffness. The system also experiences friction, given by Where b 0 is the damping coefficient. F friction = b dy dt Any other external forces such as gravitational, electrical, or magnetic forces will be lumped together as the known function F ext (t). Using Newton s Second Law, we get the second order differential equation or my = ky by + F ext (t) my + by + ky = F ext (t)

Undamped, Free Motion If we consider the simplest case, where there is no friction (b = 0) and no external forces (F ext (t) = 0) acting on the system then we get the equation m d2 y dt 2 + ky = 0 To

2 Undamped, Free Motion If we consider the simplest case, where there is no friction (b = 0) and no external forces (F ext (t) = 0) acting on the system then we get the equation m d2 y dt 2 + ky = 0 To put in standard form, we divide by m to get d 2 y dt 2 + ω2 y = 0 where ω 2 = k. This equation describes undamped free motion or simple harmonic motion m The resulting auxiliary equation is Which has complex roots Thus the general solution is Notes: r 2 + ω 2 = 0 r = ±ωi y(t) = c 1 cos ωt + c 2 sin ωt When initial conditions are used to find c 1 and c 2, 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

3 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 = c c 2 2 and φ is the phase angle defined by tan φ = c 1 c 2 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.

4 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?

5 3. Express the equation in the alternate form. y(t) = 2 3 cos 2t 1 sin 2t 6 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 m d2 y dy + b dt2 dt + ky = 0 The auxiliary equation associated with this equation is mr 2 + 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 4mk b2 β = 2m

And the general solution is y(t) = e αt (c 1 cos βt + c 2 sin βt) Or more simply y(t) = Ae αt sin(βt + φ) Where A = c 1 2 + c 2 2 and φ is the phase angle defined by tan φ = c 1 c 2 This system is

6 And the general solution is y(t) = e αt (c 1 cos βt + c 2 sin βt) Or more simply y(t) = Ae αt sin(βt + φ) Where A = c c 2 2 and φ is the phase angle defined by tan φ = c 1 c 2 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, r 1 and r 2, and a general solution y(t) = c 1 e r 1t + c 2 e r 2t This system is called overdamped because the damping force is great enough to prevent oscillation.

7 Critically Damped Motion When the discriminant is zero we get one real root, r 1, and a general solution y(t) = c 1 e r 1t + c 2 te r 1t 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.

4.2 Homogeneous Linear Equations

4.2 Homogeneous Linear Equations Homogeneous Linear Equations with Constant Coefficients Consider the first-order linear differential equation with constant coefficients a 0 and b. If f(t) = 0 then this