Mechanical Vibration System: Driving Through the Spring The ﬁgure below shows a spring-mass-dashpot system that is driven through the spring. y Spring x Mass Dashpot Figure 1. Spring-driven system Suppose that y denotes the displacement of the plunger at the top of the spring and x (t) denotes the position of the mass, arranged so that x = y when the spring is unstretched and uncompressed. There are two forces acting on the mass: the spring exerts a force given by k(y − x ) (where k is . the spring constant) and the dashpot exerts a force given by −bx (against the motion of the mass, with damping coefﬁcient b). Newton’s law gives .. . mx = k (y − x ) − bx or, putting the system on the left and the driving term on the right, .. . mx + bx + kx = ky . (1) In this example it is natural to regard y, rather than the right-hand side q = ky, as the input signal and the mass position x as the system response. Suppose that y is sinusoidal, that is, y = B1 cos(ωt). Then we expect a sinusoidal solution of the form x p = A cos(ωt − φ). By deﬁnition the gain is the ratio of the amplitude of the system response to that of the input signal. Since B1 is the amplitude of the input we have g = A/B1 . In the previous note in this session, we worked out the formulas for g and φ, and so we can now use them with the following small change. The k on the right-hand-side of equation (1) needs to be included in the gain (since we don’t include it as part of the input). We get g(ω ) = A k = = B1 | p(iω )| φ(ω ) = tan−1 bω k − mω 2 k (k − mω 2 )2 + b2 ω 2 . Note that the gain is a function of ω, i.e. g = g(ω ). Similarly, the phase lag φ = φ(ω ) is a function of ω. The entire story of the steady state system response x p = A cos(ωt − φ) to sinusoidal input signals is encoded in these two functions of ω, the gain and the phase lag. We see that choosing the input to be y instead of ky scales the gain by k and does not affect the phase lag. The factor of k in the gain does not affect the frequency where the gain is greatest, i.e. the practical resonant frequency. From the previous note in this session we know this is � k b2 ωr = − . m 2m2 Note: Another system leading to the same equation is a series RLC circuit. We will favor the mechanical system notation, but it is interesting to note the mathematics is exactly the same for both systems.