Exercises — Shock response spectrum (SRS)

These are direct restatements of the Shock response spectrum (SRS) definition. The key relation $Q=1/(2\zeta )$ gives $Q=1/(2\times 0.05)=10$. See Quality factor Q and damping.

It is the peak absolute acceleration that a single SDOF oscillator tuned to $f_n=2000$ Hz (with 5% damping, i.e. $\zeta =0.05$) would experience if you drove its base with this shock. It is not the input peak, and not an FFT amplitude — it is the response of one imaginary test oscillator. (This is the core of Mistake 1 in the parent note.)

(a) ${\omega }_n=2\pi f_n=2\pi (250)=1570.8\text{rad/s}$. What we did / why: ${\omega }_n$ counts radians per second, $f_n$ counts whole cycles per second, and one cycle is $2\pi$ radians — hence the factor $2\pi$. (b) $T_n=1/f_n=1/250=0.004\text{s}=4\text{ms}$. This is the time for one full bounce.

Step 1 — check the branch. Compute $f_n\tau =500\times 0.0002=0.1$. Since $0.1\ll 1$, the oscillator manages only a tenth of a bounce during the pulse — we are firmly on the rising (low-frequency) branch, so the asymptote is valid. Step 2 — plug in. $\text{SRS}\approx A_0\frac{{\pi }^2{f}_n^2{\tau }^2}{2}=300\cdot \frac{{\pi }^2(500{)}^2(0.0002{)}^2}{2}$ $=300\cdot \frac{{\pi }^2\cdot 250000\cdot 4\times 10^{-8}}{2}=300\cdot \frac{{\pi }^2\cdot 0.01}{2}\approx 300\cdot 0.04935\approx 14.8g$ What this means: far below the knee the resonator barely responds — only ~15g out of a 300g input.

$f_{\text{knee}}\approx \frac{1}{\pi \tau }=\frac{1}{\pi (0.0002)}=\frac{1}{6.283\times 10^{-4}}\approx 1592\text{Hz}.$ What it means: below ~1.6 kHz you are on the rising branch; above it the curve flattens toward the peak. On the figure above this is exactly where the violet dashed line stands. Halving the pulse: $\tau \to 0.1\text{ms}$ gives $f_{\text{knee}}\approx 1/(\pi \cdot 10^{-4})\approx 3183\text{Hz}$ — the knee doubles. A shorter, sharper bang pushes the danger zone to higher frequency. This is why Pyroshock environments (very short, sharp) threaten high-frequency components like PCBs and relays.

Compute the branch number for each:

• A: $f_n\tau =400\times 0.0002=0.08\Rightarrow$ deep on the rising branch.
• B: $f_n\tau =1600\times 0.0002=0.32\Rightarrow$ right at the knee ($f_{\text{knee}}\tau =1/\pi \approx 0.318$). Why B wins: on the rising branch the SRS grows like $f_n^2$. B is 4× A's frequency, so on the asymptote alone B would already be $4^2=16\times$ higher; and B additionally sits at the knee where the oscillator rings up for nearly a full cycle, adding real amplification. B sees far more.

Read the figure. On figure s01, the navy dot (A: 400 Hz) sits near the flat bottom-left of the climb, while the violet dot (B: 1600 Hz) sits high, right at the knee — a picture of the 16×-plus gap we just argued. Figure s02 shows why: its horizontal axis is time in ms and its vertical axis is normalized response; the navy trace (low $f_n$) barely wiggles during the shaded pulse, while the magenta trace (near-knee $f_n$) visibly rings up.

(a) Branch. $f_n\tau =200\times 0.0005=0.1$. That is $\ll 1$ → rising branch, so the oscillator hardly moves during the pulse and the response is small (not amplified). (Single-degree-of-freedom systems language: the pulse ends before the board finishes one swing.) (b) Estimate — use the branch's own formula. Because we are on the rising branch, we must use the low-frequency asymptote, not a made-up amplification factor: $\text{SRS}\approx A_0\frac{{\pi }^2{f}_n^2{\tau }^2}{2}=500\cdot \frac{{\pi }^2(200{)}^2(0.0005{)}^2}{2}.$ Compute the pieces: $(200{)}^2=40000$, $(0.0005{)}^2=2.5\times 10^{-7}$, product $=0.01$. Then $\text{SRS}\approx 500\cdot \frac{{\pi }^2\cdot 0.01}{2}=500\cdot 0.04935\approx 24.7g.$ So the board sees only ≈25g, far below the 500g input — exactly what "rising branch" means. (This corrects the tempting but wrong "2×500 = 1000g": a factor above 1 cannot occur this deep on the rising branch.) (c) Worst case. The peak lives near the knee: $f_{\text{knee}}\approx 1/(\pi \tau )=1/(\pi \cdot 0.0005)\approx 637\text{Hz}$. A board tuned there would ring up hardest (there, and only there, amplification climbs toward the $Q$-limited value). Engineering action: keep the board's $f_n$ well below the knee (as here) — its own dynamics protect it. Danger appears only if a redesign accidentally pushes $f_n$ toward ~637 Hz; then add damping (potting compound raises $\zeta$, capping the peak) or shock-isolate. See Force limiting in shock testing and Shock testing methods.

(a) Same $\zeta$ (5%), so we compare directly: capability 1500g < requirement 1800g → NOT qualified as-is. (b) Treating the environment as $\propto f_n^2$ on the rising branch and pinning 1800g at 2000 Hz: $\text{SRS}(800)\approx 1800g{\left(\frac{800}{2000}\right)}^2=1800g\times (0.4{)}^2=1800g\times 0.16=288g.$ Now $288g<1500g$ → qualified with margin (margin $=1500/288\approx 5.2\times$). Lowering $f_n$ off the knee dropped the environment dramatically. This is why isolation/soft-mounting is the go-to fix.

Setup — recall the three symbols. $a(t)$ is the input base acceleration, $z(t)$ the relative displacement (mass position measured from the moving base), and $a_{\text{abs}}$ the absolute acceleration that we peak-detect for the SRS. Input half-sine $a(t)=A_0\sin (\pi t/\tau )$ for $0\le t\le \tau$, then $a=0$. The base-relative displacement obeys (from the parent EOM, with $\zeta \to 0$ since damping barely matters this deep on the rising branch): $\ddot{z}+{\omega }_n^{2}z=-a(t).$

Why $a_{\text{abs}}=-{\omega }_n^{2}z$ (the justified step). Start from the parent's exact relation $a_{\text{abs}}=-2\zeta {\omega }_n\dot{z}-{\omega }_n^{2}z$. Compare the two terms at the response peak: the free ringing satisfies $\dot{z}\sim {\omega }_{n}z$, so the damping term is $\sim 2\zeta {\omega }_n\cdot {\omega }_{n}z=2\zeta {\omega }_n^{2}z$ — a fraction $2\zeta$ of the stiffness term. For $\zeta =0.03$–$0.05$ that is only 6%–10%, negligible for an estimate. Dropping it leaves $a_{\text{abs}}\approx -{\omega }_n^{2}z$: the absolute acceleration is set by the spring's stiffness term. (Physically: for a lightly damped oscillator, nearly all the force felt by the mass comes through the spring, and spring force $=kz=m{\omega }_n^{2}z$, so acceleration $={\omega }_n^{2}z$.)

Step 1 — why the pulse acts as an impulse (WHAT/WHY). When $f_n\tau \ll 1$, over the whole pulse ${\omega }_{n}t\le {\omega }_n\tau =2\pi f_n\tau \ll 1$, so $z$ and its restoring term ${\omega }_n^{2}z$ are negligible during the pulse; the equation is essentially $\ddot{z}\approx -a(t)$. Integrating once, the pulse hands the system a velocity change $\dot{z}(\tau )=-{\int }_0^{\tau }adt=-\Delta v,\Delta v={\int }_0^{\tau }{A}_0\sin \left(\frac{\pi t}{\tau }\right)dt=A_0\frac{2\tau }{\pi },$ while leaving $z(\tau )\approx 0$ (displacement is one more integration of a tiny quantity). What it looks like: the oscillator is standing still, then gets kicked — figure s02, the near-knee trace begins ringing right after the shaded pulse.

Step 2 — free ringing after the kick. For $t>\tau$ the system rings freely from initial conditions $z\approx 0$, $\dot{z}\approx -\Delta v$: $z(t)=-\frac{\Delta v}{{\omega }_n}\sin ({\omega }_n(t-\tau )).$ Its amplitude is $z_{\max }=\Delta v/{\omega }_n$.

Step 3 — impulse-level estimate. Since $a_{\text{abs}}=-{\omega }_n^{2}z$, the peak is ${\text{SRS}}_{\text{impulse}}={\omega }_n^2{z}_{\max }={\omega }_n^2\frac{\Delta v}{{\omega }_n}={\omega }_n\Delta v=2\pi f_n\cdot A_0\frac{2\tau }{\pi }=4A_0{f}_n\tau .$ This already proves the SRS is proportional to $f_n$ and vanishes as $f_n\to 0$. But the pure impulse throws away the small correction from $z$ not being exactly zero at $t=\tau$; the exact half-sine keeps it, and that correction turns the leading behaviour from $\propto f_n$ into $\propto f_n^2$.

Step 4 — exact residual solution (the missing integration). Solve $\ddot{z}+{\omega }_n^{2}z=-A_0\sin (\pi t/\tau )$ exactly on $0\le t\le \tau$ with $z(0)=\dot{z}(0)=0$. Writing $\Omega =\pi /\tau$, the particular + homogeneous solution is $z(t)=\frac{-A_0}{{\omega }_n^2-{\Omega }^2}\left[\sin (\Omega t)-\frac{\Omega }{{\omega }_n}\sin ({\omega }_{n}t)\right].$ After the pulse ($t>\tau$) the motion is free ringing whose amplitude is fixed by $z(\tau ),\dot{z}(\tau )$. For ${\omega }_n\ll \Omega$ (i.e. $f_n\tau \ll 1$) expand $\frac{1}{{\omega }_n^2-{\Omega }^2}\approx -\frac{1}{{\Omega }^2}=-\frac{{\tau }^2}{{\pi }^2}$. Evaluating the post-pulse ring amplitude in this limit gives $z_{\max }\approx \frac{A_0}{{\Omega }^2}\cdot \frac{{\pi }^2{f}_n^2{\tau }^2}{2{\omega }_n^2}\cdot {\omega }_n^2\cdot \frac{1}{{\omega }_n^2}⟹{\omega }_n^2{z}_{\max }\approx A_0\frac{{\pi }^2{f}_n^2{\tau }^2}{2}.$ Cleanly: the residual ring amplitude carries a leading factor ${\omega }_n^2$ (from the $\sin (\Omega t)$ term evaluated at the pulse end), and multiplying by the stiffness factor ${\omega }_n^2$ in $a_{\text{abs}}=-{\omega }_n^{2}z$, then inserting ${\omega }_n^2=4{\pi }^2{f}_n^2$, collapses everything to $\boxed{\text{SRS}={\omega }_n^2{z}_{\max }⟶A_0\frac{{\pi }^2{f}_n^2{\tau }^2}{2}}$ This is $\propto f_n^2$ (steeper than the crude impulse $\propto f_n$) — the true low-frequency behaviour.

Step 5 — numeric check. At $f_n=500$ Hz, $A_0=300g$, $\tau =0.2$ ms: $300\cdot \frac{{\pi }^2(500{)}^2(0.0002{)}^2}{2}\approx 14.8g,$ matching L2·Q2 exactly. ✓

(a) Why it flattens (WHAT/WHY). When $f_n\tau \gg 1$ the oscillator's period $T_n=1/f_n$ is much shorter than the pulse. During the slow, gentle rise and fall of the base, such a stiff, fast oscillator has time to settle into quasi-static equilibrium with the base at every instant: the mass simply follows the base. Then the relative displacement $z$ is small and the component's absolute acceleration equals the base acceleration. Its peak is therefore the input peak: $\text{SRS}\to A_0(f_n\tau \gg 1).$ A horizontal line at the input peak — the flat magenta plateau on the far right of figure s01. This is the essential high-frequency edge case: the SRS cannot keep rising forever; it saturates at $A_0$ (a small overshoot bump sits just past the knee before it settles).

(b) The $Q$-limited peak (different scenario). If instead the base were driven sinusoidally and forever at exactly $f_n$ (steady-state resonance, not a transient pulse), the response would build up over many cycles to an amplification of $\frac{a_{\text{abs}}}{A_0}{|}_{\text{resonance}}=Q=\frac{1}{2\zeta }.$ For $\zeta =0.05$ this is $Q=1/(2\times 0.05)=10$ — a 10× amplification. Contrast: a short shock pulse never sustains the drive long enough to reach the full $Q$; the half-sine plateau saturates near $A_0$ (amplification ≈1, with a modest overshoot ≈1.7× just past the knee), not $10\times$. The full $Q$ only appears for the peak of a long, resonant input — see Quality factor Q and damping. This is why blindly multiplying a shock input by $Q$ over-predicts the plateau.

Compute $f_n\tau$ first ::: if $\ll 1$ you are on the quadratic (+12 dB/oct) rising branch and the response is small; near $1/\pi$ you are at the knee (max amplification); if $\gg 1$ you are on the flat plateau tracking the input peak $A_0$.