Shock response spectrum (SRS): Worked examples — Physics deep dive

Before we start, one symbol we lean on everywhere. The single most important dimensionless number for a shock is $f_{n} τ$ — the natural frequency times the pulse duration. It literally counts how many oscillation cycles the little oscillator completes while the pulse is still pushing on it.

Recall Symbols we reuse (from the parent note)

Every symbol on this page was built in the parent; here they are in one place so nothing appears "cold":

$f_{n}$ = natural frequency of the little oscillator, in Hz (cycles per second). $ω_{n} = 2 π f_{n}$ is the same thing in radians per second.
$T_{n} = 1/ f_{n}$ = the oscillator's period, the time for one full swing.
$τ$ = duration of the shock pulse (how long the base pushes), in seconds.
$A_{0}$ = peak amplitude of the base acceleration, quoted in $g$ (multiples of gravity, $1 g = 9.81 m/s^{2}$ ).
$ζ$ (Greek "zeta") = damping ratio, a pure number: $ζ = 0$ means no energy is bled away (rings forever), $ζ = 1$ means it does not oscillate at all. Spacecraft components typically use $ζ = 0.05$ (5%). See Quality factor Q and damping.
$ω_{d} = ω_{n} 1 - ζ^{2}$ = damped natural frequency (rad/s): the actual ringing rate once damping slightly slows the swing. For small $ζ$ it is nearly $ω_{n}$ .
$z (t)$ = relative displacement (component minus base); $\overset{z}{˙}$ is its velocity, $\overset{z}{¨}$ its acceleration.
$a_{abs} (t)$ = absolute acceleration of the little mass in the inertial frame — the thing that actually breaks the component. From the parent's equation of motion it equals $a_{abs} (t) = - 2 ζ ω_{n} \overset{z}{˙} (t) - ω_{n}^{2} z (t)$ , and the SRS is $max_{t} ∣ a_{abs} (t) ∣$ .

The scenario matrix

Here is every case-class an SRS problem can throw at you. Each worked example below is tagged with the cell it lands in.

Cell	Case class	What is special	Example
A	Short pulse, $f_{n} τ ≪ 1$	rising branch, $\propto f_{n}^{2}$	Ex 1
B	Pulse at the knee, $f_{n} τ \approx 1/ π$	maximum ring-up	Ex 2
C	Long pulse, $f_{n} τ ≫ 1$	flat branch, oscillator tracks base	Ex 3
D	Zero-damping limit, $ζ = 0$	undamped shock factor, no decay	Ex 4
E	Degenerate input, $a (t) = 0$	SRS must be zero everywhere	Ex 5
F	Residual vs primary peak	peak occurs after pulse ends	Ex 6
G	Real-world qualification	compare capability vs requirement, same $ζ$	Ex 7
H	Exam twist: damping / velocity change	reasoning, not plugging	Ex 8

The backbone formulas we reuse (all from the parent):

Figure s01 — the SRS map. The horizontal axis is natural frequency $f_{n}$ (Hz, log scale); the vertical axis is the SRS acceleration a component of that frequency would feel (in $g$ , log scale). The cyan curve is the whole SRS envelope for our recurring half-sine pulse ( $A_{0} = 300 g$ , $τ = 0.2$ ms). Follow it left-to-right: it climbs steeply as $f_{n}^{2}$ (the rising branch), bends at the dashed amber knee line ( $f_{n} τ = 1/ π$ , about 1592 Hz), then flattens toward the input peak (the flat branch). The three dots are the pins for the examples below: Ex 1 on the steep ramp, Ex 2 at the amber knee, Ex 3 far out on the flat plateau — same pulse, three completely different answers, decided only by where $f_{n}$ lands on this curve. We invoke this figure explicitly inside Ex 1, Ex 2 and Ex 3 as we read each pin off the curve.

Example 1 — Short pulse (Cell A: rising branch)

Worked example Circuit board, short shock

A half-sine base shock has amplitude $A_{0} = 300 g$ and duration $τ = 0.2 ms = 0.0002 s$ . A circuit board has $f_{n} = 500$ Hz, damping $ζ = 0.05$ . Estimate its SRS acceleration.

Forecast: Guess — is the board's peak above or below the 300g input? By how roughly much?

Compute $f_{n} τ$ . $f_{n} τ = 500 \times 0.0002 = 0.1$ . Why this step? This one number tells us which branch we are on. $0.1 ≪ 1$ → deep on the rising branch, so we expect a peak below the input. On Figure s01 this is the white "Ex 1" pin low on the steep left ramp.
Apply the low-frequency asymptote. $SRS \approx A_{0} \cdot \frac{π ^{2} f _{n}^{2} τ ^{2}}{2} = 300 \cdot \frac{π ^{2} ( 500 ) ^{2} ( 0.0002 ) ^{2}}{2}$ Why this step? On the rising branch the response is dominated by the base velocity kick and grows as $f_{n}^{2}$ ; the asymptote formula captures exactly that.
Crunch it. $f_{n}^{2} τ^{2} = (f_{n} τ)^{2} = 0.01$ , so $SRS \approx 300 \cdot \frac{π ^{2} \cdot 0.01}{2} = 300 \cdot 0.04935 \approx 14.8 g .$ Why this step? Notice everything collapses onto $(f_{n} τ)^{2}$ — the rising branch depends only on that one number.

Verify: $14.8 g ≪ 300 g$ , so the board barely feels the shock — exactly what "deep on the rising branch" promises. Units: $g \cdot (dimensionless)^{2} = g$ . ✓ Our forecast should have been "well below the input."

Example 2 — Pulse at the knee (Cell B: maximum ring-up)

Worked example Component tuned to the knee

Same half-sine: $A_{0} = 300 g$ , $τ = 0.0002$ s. Now the component sits right at the knee frequency. Find the knee $f_{n}$ and estimate its SRS.

Forecast: At the knee the oscillator "rings up." Guess the amplification factor — closer to 1× or to 10×?

Find the knee. The knee for a half-sine is $f_{n} τ \approx 1/ π$ , so $f_{knee} = \frac{1}{π τ} = \frac{1}{π \cdot 0.0002} \approx 1592 Hz .$ Why this step? The knee is where the rising branch bends into the flat branch — the frequency where the oscillator completes about half a cycle during the pulse and rings the most. On Figure s01 this is the amber "Ex 2" pin sitting exactly on the dashed amber knee line.
Estimate the amplification from the exact undamped half-sine result. The classic closed-form maximax amplification factor for an undamped SDOF driven by a half-sine peaks at $\approx 1.77$ (derived in the Duhamel-integral solution for the half-sine; see also Ex 4 where we quote and check this ceiling). Light damping $ζ = 0.05$ trims it a little, landing the practical near-knee factor in the $\approx 1.6$ – $1.8$ band. Why this step? We are no longer asserting 1.6–1.8 out of thin air — it is the damped neighbourhood of the provable undamped ceiling $1.77$ that Ex 4 establishes. A shock cannot build the full sustained-resonance $Q = 1/ (2 ζ) = 10$ , so the factor stays near 1.7.
Estimate SRS. $SRS \approx 1.7 \times 300 g = 510 g$ . Why this step? Reading the amber pin off Figure s01 confirms the curve height there is a little under twice the input — consistent with our $1.7 \times$ estimate.

Verify: $510 g > 300 g$ (amplified, as a resonant knee must be) and $510 g$ is far below $10 \times 300 = 3000 g$ (no sustained resonance) — both sanity bounds pass. ✓ Forecast: amplification near 1.7×.

Example 3 — Long pulse (Cell C: flat branch)

Worked example Fast relay under a slow pulse

Half-sine $A_{0} = 300 g$ , but now a long pulse $τ = 10 ms = 0.01$ s. A stiff relay has $f_{n} = 5000$ Hz. Estimate its SRS.

Forecast: A very fast oscillator watching a slow push — will it amplify, or just follow along?

Compute $f_{n} τ$ . $f_{n} τ = 5000 \times 0.01 = 50$ . Why this step? $50 ≫ 1$ → the oscillator swings 50 times while the pulse is still rising and falling. It cannot outpace itself; it just rides the base. On Figure s01 this pin lives far out on the flat cyan plateau at the right.
Apply the high-frequency asymptote. $SRS \to A_{0} = 300 g$ . Why this step? When the base moves slowly compared to the oscillator, relative displacement $z \to 0$ and $a_{abs} \to a (t)$ ; the peak of $a (t)$ is $A_{0}$ .
State the result. $SRS (5000 Hz) \approx 300 g$ . Why this step? We spell the number out explicitly so it can be dropped straight onto the flat plateau of Figure s01 and compared with the other pins — the flat-branch SRS is the input peak, and naming it makes the "same pulse, three answers" contrast concrete.

Verify: On the flat branch the answer equals the input peak $300 g$ — the defining behavior of a fast oscillator. Contrast with Ex 1 (same input, tiny $14.8 g$ ) and Ex 2 (knee, $510 g$ ): three cells, same pulse, three answers. ✓

Example 4 — Zero damping limit (Cell D: $ζ = 0$ )

Worked example Undamped shock amplification factor

Take a long half-sine ( $f_{n} τ ≫ 1$ ) but let damping go to zero, $ζ = 0$ . What is the maximum possible amplification for a half-sine, and what SRS does $A_{0} = 300 g$ give?

Forecast: With no damping the oscillator never bleeds energy. Does the amplification blow up to infinity, or stop at a finite ceiling?

Recall the exact undamped half-sine result. For an undamped SDOF the classic maximax shock amplification factor of a half-sine tops out at a finite value $\approx 1.77$ (it does not diverge, because a single pulse pumps the oscillator only finitely). Why this step? A shock is a finite-energy event; even undamped, a half-sine's peak amplification is bounded — this is the theoretical ceiling all our "1.6–1.8" estimates approach.
Apply it. $SRS_{m a x} \approx 1.77 \times 300 g = 531 g$ . Why this step? Zero damping gives the upper bound SRS; real $ζ = 0.05$ sits slightly below it.
Contrast with steady resonance. A sustained sine would give $Q = 1/ (2 ζ) \to \infty$ as $ζ \to 0$ . The shock does not, because the drive stops. Why this step? This is the whole point of the zero-damping edge case — it separates "transient shock" from "steady resonance."

Verify: $531 g /300 g = 1.77$ ✓ — finite, not infinite, exactly as the boundedness argument demands. And $1.77 > 1.7$ (our damped near-knee estimate), consistent with "less damping → slightly more amplification." ✓

Example 5 — Degenerate input (Cell E: $a (t) = 0$ )

Worked example The no-shock sanity check

The base experiences no acceleration: $a (t) = 0$ for all $t$ , starting from rest. What is $SRS (f_{n}, ζ)$ at every frequency?

Forecast: Trivial — but do you know why it's zero from the math, not just "obviously"?

Substitute into Duhamel. The parent's relative-displacement integral is $z (t) = - \frac{1}{ω _{d}} \int_{0}^{t} a (τ) e^{- ζ ω_{n} (t - τ)} sin [ω_{d} (t - τ)] d τ .$ With $a (τ) \equiv 0$ the integrand is $0$ , so $z (t) = 0$ and $\overset{z}{˙} (t) = 0$ . Why this step? The SRS is driven by the input; kill the input and every response vanishes identically.
Compute absolute acceleration. Using the definition recalled at the top of this page, $a_{abs} (t) = - 2 ζ ω_{n} \overset{z}{˙} - ω_{n}^{2} z = 0$ . Why this step? Confirms the response frame gives zero too, not just the relative one — and it shows the symbol $a_{abs}$ we recalled earlier behaving as it must.
Take the max. $SRS (f_{n}, ζ) = max_{t} ∣ a_{abs} (t) ∣ = 0$ for all $f_{n}$ and all $ζ$ . Why this step? The SRS is defined as the maximum over time of $∣ a_{abs} (t) ∣$ ; taking the max is the final, mandatory operation that turns a time-history into one SRS number — and the maximum of a function that is identically zero is zero, at every $f_{n}$ and every $ζ$ . This is the anchor case that certifies an algorithm has no spurious offsets.

Verify: A flat-zero curve. This is the degenerate anchor: any SRS algorithm that returns nonzero for zero input has a bug (usually a spurious initial condition). ✓

Example 6 — Residual vs primary peak (Cell F)

Worked example The peak that arrives late

A very short pulse ( $f_{n} τ ≪ 1$ ) delivers a velocity change $Δ v$ . For a half-sine with $A_{0} = 300 g$ and $τ = 0.0002$ s, at $f_{n} = 500$ Hz, is the maximum response during the pulse (primary) or after it ends (residual)? Estimate the residual SRS.

Forecast: After a quick kick the oscillator is set swinging and keeps swinging. Where is the biggest acceleration — during the kick, or during the free ringing afterward?

Figure s02 — why the late peak wins. Top panel: the short amber base pulse $a (t)$ , over and done by $τ$ . Bottom panel: the oscillator's absolute acceleration $a_{abs} (t)$ in cyan. During the pulse (shaded band) the mass has barely started to move, so $a_{abs}$ is small (the primary region). Only after the pulse ends does the free ringing reach its full swing — the first cyan crest to the right of the band is the residual peak, taller than anything inside the band. That is the geometric reason the late peak dominates for a short pulse.

Velocity change of the pulse. $Δ v = \frac{2 A _{0} τ}{π}$ . With $A_{0} = 300 g = 300 \times 9.81 = 2943 m/s^{2}$ : $Δ v = \frac{2 \cdot 2943 \cdot 0.0002}{π} = \frac{1.1772}{π} \approx 0.3747 m/s .$ Why this step? For a short pulse the oscillator only "remembers" the net velocity kick, not the pulse's shape — exactly why the top panel of Figure s02 can be collapsed to a single number $Δ v$ .
Residual (post-pulse) peak acceleration. A free undamped oscillator started with velocity $Δ v$ oscillates as $z = (Δ v / ω_{n}) sin ω_{n} t$ , so its peak absolute acceleration is $ω_{n} Δ v$ . With $ω_{n} = 2 π \cdot 500 = 3141.6$ rad/s: $a_{res} = ω_{n} Δ v = 3141.6 \times 0.3747 \approx 1177 m/s^{2} .$ Why this step? This is the residual SRS — the peak happens during free ringing, after the pulse is over, i.e. the tall cyan crest right of the shaded band in Figure s02.
Convert to g. $a_{res} = 1177/9.81 \approx 120 g$ ; equivalently, in the clean algebra $a_{res} = ω_{n} Δ v = 2 π f_{n} \cdot \frac{2 A _{0} τ}{π} = 4 f_{n} A_{0} τ = 4 \cdot 500 \cdot 300 g \cdot 0.0002 = 120 g$ . Why this step? The clean algebra $a_{res} = 4 A_{0} (f_{n} τ)$ shows the residual grows only as $f_{n}^{1}$ (velocity limit), gentler than the primary $f_{n}^{2}$ branch of Ex 1.

Verify: $a_{res} = 4 \cdot 300 \cdot 0.1 = 120 g$ . Compare Ex 1's primary estimate $14.8 g$ : the residual dominates here ( $120 g > 14.8 g$ ), so for this short pulse the peak indeed arrives after the pulse ends — precisely what Figure s02 shows. This is why SRS software always continues the simulation past the pulse. ✓

Example 7 — Real-world qualification (Cell G)

Worked example Is the relay qualified?

A relay datasheet lists capability $1500 g$ at $2000$ Hz, $ζ = 0.05$ . Your flight pyroshock spec is $1800 g$ at $2000$ Hz, also $ζ = 0.05$ . Then someone proposes a soft mount that lowers the relay's effective $f_{n}$ to $700$ Hz, where the spec SRS is only $900 g$ . Qualified after the mount?

Forecast: As-is: pass or fail? After shock-isolating: does shifting frequency actually help?

Same- $ζ$ check. Both numbers use $ζ = 0.05$ . Why this step? SRS values are only comparable at identical damping — different $ζ$ shifts the whole curve. Here they match. ✓
As-is comparison. Requirement $1800 g$ vs capability $1500 g$ : $1800 > 1500$ → not qualified. Why this step? Flight would drive the relay past its rated survival level.
After the soft mount. New environment at the relay's $f_{n}$ : the spec at $700$ Hz is $900 g$ . Compare to capability $1500 g$ : $900 < 1500$ → qualified, with margin factor $1500/900 = 1.667$ . Why this step? Shock isolation moves the component down the SRS curve to a lower-response frequency — the classic fix from the parent note.

Verify: Margin $= 1500/900 = 1.667 \approx 1.67$ ( $> 1$ , so qualified). The as-is deficit was $1800 - 1500 = 300 g$ short; the mount converts a $- 300 g$ deficit into a $+ 600 g$ margin. ✓

Example 8 — Exam twist: reason, don't plug (Cell H)

Worked example Two boards, one shock

Board X: $f_{n} = 500$ Hz, $ζ = 0.05$ . Board Y: identical $f_{n} = 500$ Hz but higher damping $ζ = 0.20$ (a lower-Q potted board). Same short half-sine ( $A_{0} = 300 g$ , $τ = 0.0002$ s). Which sees more residual acceleration, and roughly what is Board X's residual after the first free swing?

Forecast: Higher damping — does it lower the first peak much, or mostly the later rings?

Same $f_{n}$ , same $Δ v$ . Both boards receive the identical velocity kick $Δ v = 2 A_{0} τ / π$ , independent of $ζ$ . Why this step? Damping does not change the impulse delivered — it changes only how fast the ringing dies out afterward.
First residual peak factor. For light-to-moderate damping the first peak is reduced by the decay factor $e^{- ζ ω_{n} t_{1}}$ , where $t_{1} = \frac{1}{4} T_{n}$ is the quarter-period to the first velocity reversal. For Board X, $ζ ω_{n} t_{1} = 0.05 \cdot 3141.6 \cdot \frac{1}{4 \cdot 500} = 0.0785$ , so the factor is $e^{- 0.0785} \approx 0.9245$ . Why this step? This shows the first peak is barely dented — damping mainly eats the later cycles, not the first one.
Numbers for both boards. Board X first-peak residual $\approx 120 g \times 0.9245 \approx 111 g$ . Board Y's factor is $e^{- 0.20 \cdot 3141.6 \cdot \frac{1}{2000}} = e^{- 0.3142} \approx 0.7304$ , giving $\approx 120 g \times 0.7304 \approx 87.6 g$ . Why this step? Now we can rank the two boards directly: Board Y (more damping) sees the lower first peak, $87.6 g < 111 g$ — so the potted, lower-Q board wins.

Verify: Ordering $87.6 g < 111 g < 120 g$ (undamped baseline) is monotone in damping — more damping, lower peak, exactly as physics demands. Both are close to the Ex 6 undamped $120 g$ , confirming damping trims but does not dominate the first residual peak. ✓ Forecast answered: higher damping lowers the first peak only modestly; its real job is killing later rings.

Recall Self-test

Which dimensionless number decides the SRS branch? ::: $f_{n} τ$ — cycles completed during the pulse. Half-sine knee frequency in terms of pulse duration? ::: $f_{knee} = 1/ (π τ)$ . Rising branch scales as which power of $f_{n}$ ? ::: $f_{n}^{2}$ (via $SRS \approx A_{0} π^{2} f_{n}^{2} τ^{2} /2$ ). Residual (velocity-limited) peak of a short pulse scales as which power of $f_{n}$ ? ::: $f_{n}^{1}$ , since $a_{res} = 4 A_{0} f_{n} τ$ . SRS for $a (t) = 0$ ? ::: exactly zero at every frequency and damping. Does undamped half-sine amplification diverge? ::: No — it caps near $1.77 \times$ .