Shock response spectrum (SRS)

What Is the Shock Response Spectrum?

Why SDOF oscillators? Every real component (circuit board, fuel tank mount, antenna bracket) can be approximated as a mass-spring-damper near its fundamental resonance. The SRS pre-computes the worst-case response for all possible resonances.

Deriving the SRS from First Principles

Step 1: Equation of Motion for a Single Oscillator

Consider a component with mass $m$ , stiffness $k$ , damping $c$ , attached to a base experiencing acceleration $a(t)$ . Let $z(t)$ be the relative displacement (component position minus base position).

Free-body diagram gives: $m\ddot{z} + c\dot{z} + kz = -ma(t)$

Why the minus sign? The base acceleration $a(t)$ is an inertial force acting opposite to the base motion direction. Think of yourself in an accelerating car—you feel pushed backward.

Divide by $m$ : $\ddot{z} + 2\zeta\omega_n\dot{z} + \omega_n^2 z = -a(t)$

where:

$\omega_n = \sqrt{k/m}$ is the natural frequency (rad/s)
$\zeta = c/(2\sqrt{km})$ is the damping ratio
$f_n = \omega_n/(2\pi)$ is natural frequency in Hz

Step 2: Duhamel's Integral (Convolution Solution)

For an underdamped system ( $\zeta < 1$ ), the relative displacement response is:

$z(t) = -\frac{1}{\omega_d}\int_0^t a(\tau) e^{-\zeta\omega_n(t-\tau)} \sin[\omega_d(t-\tau)] \, d\tau$

where $\omega_d = \omega_n\sqrt{1-\zeta^2}$ is the damped natural frequency.

Why this form? Each impulse $a(\tau)d\tau$ at time $\tau$ creates a decaying oscillation; we sum (integrate) all contributions from $\tau=0$ to $t$ .

Step 3: Absolute Acceleration

The absolute acceleration of the component (what actually breaks things) is: $a_{\text{abs}}(t) = a(t) + \ddot{z}(t)$

Why add $a(t)$ ? The relative acceleration $\ddot{z}$ is measured from the moving base; we need total acceleration in the inertial frame.

From the EOM: $\ddot{z} = -a(t) - 2\zeta\omega_n\dot{z} - \omega_n^2 z$ , so: $a_{\text{abs}}(t) = -2\zeta\omega_n\dot{z}(t) - \omega_n^2 z(t)$

Step 4: Maximum Absolute Acceleration

The SRS value at natural frequency $f_n$ and damping $\zeta$ is: $\text{SRS}(f_n, \zeta) = \max_{t \geq 0} |a_{\text{abs}}(t)|$

Computational procedure:

Pick a natural frequency $f_n$ (e.g., 100 Hz)
Numerically integrate the Duhamel convolution to get $z(t)$ and $\dot{z}(t)$
Compute $a_{\text{abs}}(t) = -2\zeta\omega_n\dot{z} - \omega_n^2 z$
Find the peak: $\max|a_{\text{abs}}(t)|$
Repeat for 100–1000 frequencies from ~10 Hz to 10,000 Hz
Plot the envelope

Physical Interpretation

Reading the SRS curve:

Low frequencies (10–100 Hz): Large structural modes; here the oscillator period is much longer than the pulse, so the SRS rises steeply with frequency (approximately $+40$ dB/decade, i.e. $+12$ dB/octave, proportional to $f_n^2$ )
Mid frequencies (100–1000 Hz): Secondary structure (panels, brackets); the knee and peak amplification often lie here for pyroshock
High frequencies (1000–10,000 Hz): Component-level resonances (PCBs, relays); beyond the knee a simple pulse SRS flattens toward the peak acceleration (≈ $0$ dB/octave), because these fast oscillators simply track the base peak
Knee frequency ≈ $1/(\pi\tau)$ for a half-sine of duration $\tau$ ; a shorter pulse pushes the knee to higher frequency

Worked example Worked Example 1: Stage Separation Shock

Given: Explosive bolt firing creates a 500g peak shock pulse lasting 0.5 ms. A circuit board has $f_n = 200$ Hz, $\zeta = 0.03$ . Find: Will the board see more than 1000g?

Solution:

Compare pulse duration to oscillator period. For $f_n = 200$ Hz, $T_n = 1/200 = 5$ ms. So $0.5$ ms $\approx T_n/10$ — the pulse is much shorter than one oscillation period. Why this step? When the pulse is short relative to $T_n$ , the oscillator only "sees" an impulse; but at $f_n\tau = 200 \times 0.0005 = 0.1$ we are still on the rising (low-frequency) part of the SRS, so amplification is modest.
Estimate the response. On the rising branch the SRS scales roughly with $f_n$ times the velocity change. A typical amplification factor here is ≈ 1.5–2× for low damping (not the full resonant 5–10× that occurs only near the knee).
Estimate: $\text{SRS}(200 \text{ Hz}) \approx 2 \times 500g = 1000g$ — right at the limit; marginal.
Why this matters? The board sits on the rising part of the SRS. If instead $f_n$ were near the knee ( $\approx 1/(\pi\tau) \approx 640$ Hz), it would see much higher amplification. Engineering action: Keep $f_n$ low (away from the knee) or add damping (potting compound) to reduce peak response.

Worked example Worked Example 3: Estimating SRS from Pulse Shape

Idealized half-sine pulse: Duration $\tau = 0.2$ ms $= 0.0002$ s, amplitude $A_0 = 300g$ .

Approximate SRS at $f_n = 2000$ Hz:

Compute the key dimensionless number: $f_n\tau = 2000 \times 0.0002 = 0.4$ . Why this step? $f_n\tau$ decides which branch of the SRS we are on—well below 1 means low-frequency (rising) branch, near/above 1 means we are approaching or past the knee.
The low-frequency asymptote (valid when $f_n\tau \ll 1$ ) is $\text{SRS} \approx A_0 \cdot \dfrac{\pi^2 f_n^2 \tau^2}{2}$ , which rises as $f_n^2$ (i.e. $+40$ dB/decade). This is the low-frequency behavior, not the high-frequency one.
Here $f_n\tau = 0.4$ is approaching the knee ( $f_n\tau \approx 1/\pi \approx 0.32$ for the knee), so the simple $f_n^2$ formula is beginning to break down and the curve is transitioning toward its flat, peak-amplification region.
Practical estimate near the knee: amplification ≈ 1.6–1.8× → $\text{SRS}(2000 \text{ Hz}) \approx 1.8 \times 300g \approx 540g$ .
Why this step? The oscillator at 2000 Hz completes just under one cycle during the pulse, so it partially "rings up" — the classic knee behavior between the rising branch and the flat high-frequency branch.

Common Mistakes

Why Engineers Use SRS Instead of Raw Time Histories

Three reasons:

Universality: One SRS curve characterizes the shock for all possible component resonances. Testing to the SRS envelope qualifies every part, regardless of its specific $f_n$ .
Comparability: You can overlay SRS from different events (launch, stage sep, docking) and take the worst-case envelope as the design spec.
Test reproducibility: Pyroshock time histories are chaotic and unrepeatable. But you can generate a synthetic shock pulse (via shaker or pyro simulator) that matches the SRS—achieving the same damage potential with different waveforms.

Analogy: SRS is to shock what a "design load factor" is to aerodynamics—a single number (or curve) that bounds the complexity of reality.

Recall Explain to a 12-Year-Old

Imagine you have 100 different bells, each ringing at a different pitch (frequency). Now you hit the table they're all sitting on with a hammer (that's your shock). Some bells will ring SUPER LOUD because the hammer hit matches their special pitch—that's called resonance. Other bells barely make a sound.

The Shock Response Spectrum is like a report card that tells you: "The bell that rings at 50 Hz got shaken to 200g, the 100 Hz bell got 500g, the 1000 Hz bell got 800g..." You get one number for every possible bell pitch.

Why is this useful? Because your spacecraft has thousands of parts, each like a different bell. Instead of testing every part separately with the hammer, you just look at the SRS curve and say, "Okay, my circuit board rings at 150 Hz, so the curve tells me it'll see 300g. Can it survive 300g? Yes? Great, we're done!"

The SRS is a cheat sheet that predicts the worst shaking for every single part from just one hammer hit.

Connections

Pyroshock environments — Origin of severe SRS specs in spacecraft
Single-degree-of-freedom systems — Foundation: every SRS point is an SDOF response
Duhamel's integral — Mathematical engine for computing transient response
Quality factor Q and damping — $Q =1/(2\zeta)$ ; relates to SRS peak sharpness
Mechanical impedance — SRS ties to how components "load" the base structure
Modal analysis — Real structure = superposition of SDOF modes; SRS tests each
Shock testing methods — How to reproduce SRS in the lab (drop tables, resonant plates)
Vibration power spectral density (PSD) — SRS for shocks, PSD for random vibration
MIL-STD-810 Method 516 — Military shock test standard; specifies SRS test profiles
Acceleration response in structures — SRS is the worst-case envelope of this
Force limiting in shock testing — Prevents over-testing at high frequencies

#flashcards/physics

What does each point on an SRS curve represent? :: The maximum absolute acceleration experienced by a single-degree-of-freedom (SDOF) oscillator with that natural frequency and specified damping, when subjected to the shock input.

Why is SRS preferred over raw time histories for spacecraft shock specs?

SRS provides a universal envelope covering all possible component resonances, enables comparison between different shock sources, and allows reproducible testing with synthetic shocks that match the damage potential.

Write the equation of motion for an SDOF system under base excitation

a(t)

\ddot{z} + 2\zeta\omega_n\dot{z} + \omega_n^2 z = -a(t)

, where

z

is relative displacement and the minus sign accounts for inertial force direction.

How is absolute acceleration related to relative motion in SRS analysis?

a_{\text{abs}}(t) = a(t) + \ddot{z}(t) = -2\zeta\omega_n\dot{z}(t) - \omega_n^2 z(t)

, combining base acceleration with component's acceleration relative to the base.

What damping ratio is standard for spacecraft SRS specifications?

\zeta = 0.05

(5%), also called Q=10, representing typical lightly-damped aerospace structures.

For a half-sine shock pulse of duration

\tau

, roughly where is the SRS knee?

Near

f_n \approx 1/(\pi\tau)

; below the knee the SRS rises as

f_n^2

(+40 dB/decade), above the knee it flattens toward the peak acceleration.

What is the low-frequency asymptotic behavior of the SRS for a short pulse?

It rises as

f_n^2

(approximately

+40

dB/decade, or

+12

dB/octave), described by

\text{SRS}\approx A_0\,\pi^2 f_n^2\tau^2/2

when

f_n\tau\ll1

What is the high-frequency behavior of a simple-pulse SRS beyond the knee?

It flattens (≈ 0 dB/octave) toward the peak base acceleration, because very fast oscillators simply track the base peak.

What is the difference between SRS and FFT of a shock pulse?

FFT shows the frequency content (energy distribution) of the input signal itself; SRS shows the maximum response amplification for resonant systems at each frequency.

Why must SRS specifications always include damping ratio?

SRS magnitude depends strongly on damping—5% damping gives 2-3× higher peaks than 10% damping—so comparisons require matched damping assumptions.

How does SRS relate to actual component natural frequencies?

Each component resonates like an SDOF system; the SRS value at the component's natural frequency predicts its peak acceleration, guiding design and qualification.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Shock Response Spectrum (SRS) kya hai aur kyun zaroori hai?

Jab spacecraft mein explosive bolt fire hota h

What Is the Shock Response Spectrum?

Deriving the SRS from First Principles

Step 1: Equation of Motion for a Single Oscillator

Step 2: Duhamel's Integral (Convolution Solution)

Step 3: Absolute Acceleration

Step 4: Maximum Absolute Acceleration

Physical Interpretation

Common Mistakes

Why Engineers Use SRS Instead of Raw Time Histories

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Spacecraft Structures & Systems Engineering

Connections