# A mass on a spring, three ways Hang a spring of natural length $l$ from the ceiling. Hang a mass $m$ on it and it stretches to a new rest position, a distance $L$ further down. Pull the mass down, let go, and measure its displacement $y(t)$ from that rest position, with down counted as positive. Newton's second law, with a dashpot of strength $b$ and an applied force $F(t)$, says $ m\,y'' = mg - b\,y' - k\,(L + y) + F(t). $ At rest nothing moves, so $mg = kL$, and those two terms cancel. What is left is the equation for the rest of the unit, $ m\,y'' + b\,y' + k\,y = F(t). $ The three applets below take this apart in order. The first is the derivation itself. The second is the free motion, $F = 0$. The third turns the force on. > [!info] Reading the applets > In every applet the graph of $y$ uses a **downward** vertical axis at the same scale as the picture of the mass, so the mass and its point on the graph sit at the same height. You can drag the mass (in the first two) and let go. ## 1. Where gravity goes <div class="applet" data-applet="spring-hang"></div> The three columns are the bare spring, the mass hanging at rest, and the mass in motion. On the moving mass the arrows are the forces. Measured from rest, gravity $mg$ and the part $kL$ of the spring force always cancel, and only $-ky$ and $-by'$ are left to move anything. Now switch to measuring $x = L + y$ from the end of the unstretched spring. Nothing about the motion changes. The equation does, $ m\,x'' + b\,x' + k\,x = mg , $ and gravity is back as a constant forcing term whose only effect is to move the equilibrium to $x = mg/k = L$. **Try this.** - Press *hang it from the unstretched length and let go*. How far below the rest position does the mass go? Why exactly that far? - Double $m$. What happens to $L$? To the period? - The readout gives $\omega_0 = \sqrt{k/m} = \sqrt{g/L}$. Explain why you could find the period of the oscillation with a ruler and no scale. > [!question]- Reveal > Let go at $y = -L$ with no velocity and no damping, and the motion is $y = -L\cos\omega_0 t$, so the mass falls to $y = +L$, twice the static stretch below the unstretched end. Doubling $m$ doubles $L = mg/k$ and multiplies the period $2\pi\sqrt{m/k}$ by $\sqrt 2$. Since $k/m = g/L$, measuring the static stretch alone gives $\omega_0 = \sqrt{g/L}$. ## 2. Free motion and the characteristic roots <div class="applet" data-applet="spring-damper"></div> Trying $y = e^{rt}$ in $m\,y'' + b\,y' + k\,y = 0$ gives $m r^2 + b r + k = 0$, so $ r = \frac{-b \pm \sqrt{b^2 - 4mk}}{2m}. $ Everything about the motion is in where those two roots sit. The real part is the decay rate, the imaginary part is the frequency. The critical value is $b_c = 2\sqrt{mk}$, and $\zeta = b/b_c$ is the damping ratio. | | roots | motion | |---|---|---| | $b = 0$ | $\pm i\omega_0$ | oscillates forever, $\omega_0 = \sqrt{k/m}$ | | $0 < b < b_c$ | $-\tfrac{b}{2m} \pm i\mu$ | oscillates inside the envelope $e^{-bt/2m}$ | | $b = b_c$ | $-\omega_0$, twice | $y = (c_1 + c_2 t)e^{-\omega_0 t}$ | | $b > b_c$ | two negative reals | creeps home without crossing more than once | **Try this.** - Raise $b$ slowly from $0$ and watch the roots. While they are complex, what curve do they move along? Why? (Multiply the two roots together.) - Keep raising $b$ past $b_c$. One root heads left, one heads back toward $0$. Which one controls how long the mass takes to get home? - The readout reports the slowest decay rate. For which $b$ is it largest? - Release the mass from rest. Does critical damping get it within $2\%$ of rest soonest? Try $\zeta$ around $0.7$ to $0.8$. > [!question]- Reveal > The product of the roots is $k/m$, so while they are a conjugate pair $|r|^2 = k/m$ and they slide around the circle of radius $\omega_0$. Past $b_c$ the slow root $r_+ = \frac{-b + \sqrt{b^2 - 4mk}}{2m}$ moves back toward $0$ as $b$ grows, so heavy damping makes the return *slower*. The slowest decay rate is largest, equal to $\omega_0$, exactly at $b = b_c$. Critical damping is the fastest return *that never overshoots*. If a small overshoot is allowed, a little less damping wins: released from rest, the time to stay within $2\%$ is shortest near $\zeta \approx 0.78$, about $0.9/\omega_0$ against about $1.46/\omega_0$ at critical damping. ## 3. Forcing, beats and resonance <div class="applet" data-applet="spring-forced"></div> With $F(t) = F_0 \cos\omega t$ the solution is $y = y_h + y_p$. The homogeneous part $y_h$ is the free motion from section 2, and it dies away whenever $b > 0$. The particular part is the steady state, $ y_p = A\cos(\omega t - \varphi), \qquad A = \frac{F_0}{\sqrt{(k - m\omega^2)^2 + (b\omega)^2}}, \qquad \tan\varphi = \frac{b\omega}{k - m\omega^2}. $ The dashed purple box in the picture is $F(t)/k$, where the force by itself would hold the mass if it changed slowly. The lower panels plot the magnification $A\,k/F_0$ and the lag $\varphi$ against $\omega/\omega_0$, and both depend only on $\zeta$. **Try this.** - *slow drive*, then *fast drive*. Compare the mass with the purple box each time. In step, or opposite? - *at the peak*. Where is the mass when the force is largest? Read $\varphi$. - *beats*. Turn off $y_h$ and $y_p$ and look at $y$ alone. Then turn them back on. What is the beat made of? - *pure resonance*. How fast does the amplitude grow? - Sweep $\omega$ with $\zeta = 0.1$, then with $\zeta = 0.5$. When does the measured curve follow the steady-state curve, and when does it fall behind? > [!question]- Reveal > Driven slowly, $\varphi \approx 0$ and the mass follows $F/k$. Driven fast, $\varphi \approx 180^\circ$ and the mass moves opposite to the force with a small amplitude, about $F_0/(m\omega^2)$. At $\omega = \omega_0$ the lag is exactly $90^\circ$, so the force is largest when the mass passes through rest at full speed, which is when pushing does the most work. With $b = 0$, starting from rest, $y_h$ and $y_p$ have nearly equal and opposite amplitudes, and their sum is > $y = \frac{2F_0}{m(\omega_0^2 - \omega^2)} \sin\frac{(\omega_0 - \omega)t}{2}\, \sin\frac{(\omega_0 + \omega)t}{2},$ > a fast oscillation inside a slow envelope. At $\omega = \omega_0$ the steady state is $y_p = \frac{F_0}{2m\omega_0}\,t\sin\omega_0 t$, which grows linearly without bound. A sweep is only as good as the transient is short. With $\zeta = 0.5$ the measured peak is within a percent of $1/(2\zeta\sqrt{1-\zeta^2})$. With $\zeta = 0.1$ it comes out about $6\%$ low and late, near $\omega/\omega_0 \approx 1.07$, because the mass is still catching up with a frequency that has already moved on.