Free vibrations
In the introductory tutorial in this series we considered a free vibrating translational spring and mass system with one degree of freedom as shown in Figure 1 below. Mass m connected to a spring with spring constant k is free to move on a horizontal frictionless plane from the static equilibrium position x = 0. The diagram shows the force Fs exerted by the spring on mass m in extension (i) and compression (ii).
From dynamic* free body diagrams we obtained the following equation of motion in terms of displacement x and time t noting that this equation is an expression of simple harmonic motion (SHM).
* showing forces producing acceleration of mass m, not static equilibrium
We now find a function x(t) that is a general solution to this equation. If you are not familiar with solving second-order homogeneous linear differential equations using the characteristic equation method refer to the maths tutorial.
Note that k has dimensions ML / T2L (force / displacement). Thus ω2 and ω = √(k / m) have dimensions respectively 1 / T2 and 1 / T.
From the maths tutorial we see that two complex conjugate roots of the characteristic equation expressed as:
r1 =( α + iβ) and r2 = (α - iβ) give the following general solution for x = f(t):
In the case above r1, 2 = ± iω Thus α = 0 and β = ω giving the general solution for the equation of motion as:
It is conventional to replace constants C1 and C2 with A and B giving:
x(t) = [Acos(ωt) + Bsin(ωt)] ---------- (3)
We now find a particular solution for (3) using the following initial conditions at time t = 0 :
This means that at time t = 0 the extension of the spring is x0 and the mass is given initial velocity v0.
Inserting initial condition x0 into equation (3) at time t = 0 gives:
x0 = Acos(ωt) + Bsin(ωt) = (A).(1) + (B).(0) = A thus A = x0
v0 = -ωA.sin(ωt) + ωB.cos(ωt) = -ωA.(0) + ωB.(1) thus B = (v0 / ω)
x(t) can now be expressed in terms of initial conditions x0 and v0 as;
x(t) = x0cos(ωt) + (v0 / ω) sin(ωt) ------- (4)
It is convenient to express the equation of motion in terms only of cos(ωt) using the following identity:
Acos(ωt) + Bsin(ωt) = Rcos(ωt - φ) where R = √(A2 +B2) and φ = tan-1 B /A
R is the maximum absolute value (or amplitude) of x(t).
ω represents the angular frequency of the SHM in radians/sec, called the natural frequency of the system denoted by ωn . ωn / 2π is the natural frequency measured in Hertz. The period T of one cycle is the reciprocal of the frequency = 2π / ωn .
Figure 2 below plots displacement x(t) = R.cos(ωt - φ) against time t for a spring and mass system with the following parameters:
mass m = 1 kg spring constant k = 10 N/m initial displacement x0 = 0.1 m initial velocity v0 = 1 m/s.
In this example ωn = √(k / m) = √(10 / 1) = √((10) rad/sec equivalent to (ωn / 2π) = 0.503 Hz.. Period T = (2π / ωn ) = (1 / 0.503) = 1.99 s. Amplitude R = 0.332 m.
Figure 2 also plots x(t) = R.cos(ωt) which represents SHM with initial conditions x0 = R and v0 = 0. Note that amplitude and frequency of both motions are identical. The phase angle φ between the two oscillations in this example = 72.5° with R.cos(ωt - φ) lagging R.cos(ωt)
In the next tutorial we consider free vibrations with damping.
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