Shock response spectrum (SRS): Question bank — Physics deep dive

Prerequisites worth reviewing first: Single-degree-of-freedom systems, Duhamel's integral, Quality factor Q and damping, Pyroshock environments.

Symbols used on this page (defined before you meet them)

Definition The cast of symbols

Before answering a single trap, pin down every symbol, because most traps are really symbol confusions in disguise.

$a (t)$ — the base acceleration time history: how hard the support under a component is shaken, as a function of time $t$ . This is the shock "input".
$z (t)$ — the relative displacement: the position of the component's mass minus the position of its base. It measures how much the mass lags or leads the shaking base, not its position in the room.
$\overset{z}{˙}$ , $\overset{z}{¨}$ — the first and second time derivatives of $z$ : relative velocity and relative acceleration.
$ω_{n}$ — the natural angular frequency in radians per second, $ω_{n} = k / m$ for stiffness $k$ and mass $m$ : how fast the oscillator would ring if plucked and left alone.
$f_{n}$ — the natural frequency in hertz, $f_{n} = ω_{n} / (2 π)$ : the same ringing rate counted in cycles per second.
$ζ$ (zeta) — the damping ratio, a dimensionless number, $ζ = c / (2 k m)$ for damping coefficient $c$ : how quickly the ringing dies out. $ζ < 1$ under-damped (rings), $ζ = 1$ critically damped (no ring), $ζ > 1$ over-damped (crawls back slowly).
$Q$ — the quality factor, $Q = 1/ (2 ζ)$ : the peak steady-state amplification at resonance. $ζ = 0.05$ means $Q = 10$ . See Quality factor Q and damping.
$τ$ (tau) — the pulse duration: how long the shock input $a (t)$ lasts (e.g. a half-sine of width $τ$ ).

Figures that anchor the traps

The sign of the forcing term and the shape of the SRS are the two things people get wrong most. Both are settled by pictures.

Look at the base-excitation diagram above: the mass $m$ hangs on spring $k$ and damper $c$ from a base that is itself being flung by $a (t)$ . Because we track the mass relative to that accelerating base, an inertial (pseudo-)force $- m a (t)$ appears on the mass — pointing opposite to the base acceleration. That is the origin of the minus sign in $m \overset{z}{¨} + c \overset{z}{˙} + k z = - m a (t)$ .

The second figure shows a half-sine pulse of width $τ$ and, beneath it, a schematic SRS: a steep rising branch ( $\propto f_{n}^{2}$ , $+ 40$ dB/decade), a knee near $f_{n} \approx 1/ (π τ)$ , and a flat high-frequency branch that levels off at the input peak. Refer back to it whenever a trap mentions "branch", "knee", or "asymptote".

True or false — justify

The SRS is a property of the shock pulse alone, independent of the oscillators used to compute it.

False. The SRS is a response curve — it depends on the assumed damping ratio

ζ

; the same pulse gives a lower, smoother SRS at high damping and a peakier one at low damping.

The SRS and the Fourier transform (FFT) of the same pulse carry the same information.

False. The FFT describes the input's frequency content; the SRS describes how a family of resonators respond. A short pulse has broadband FFT but an SRS with a distinct knee.

A power spectral density (PSD) is just another name for the FFT.

False. A PSD is the squared magnitude of the FFT divided by bandwidth (units of

g^{2} / Hz

), so it discards phase and expresses power per unit frequency; the raw FFT keeps amplitude and phase. Neither one is the SRS.

Two completely different-looking time histories can have identical SRS curves.

True. Many pulse shapes with the same net velocity change

Δ v = \int a d t

and similar duration collapse onto nearly the same SRS — the SRS discards phase and much waveform detail, which is why SRS specs alone don't uniquely define a test signal.

At very high natural frequency the SRS of a short pulse approaches the peak of the input acceleration

a (t)

True. A very stiff/fast oscillator has a period far shorter than the pulse, so it rigidly tracks the base — its absolute acceleration equals the base acceleration, whose maximum is the input peak.

Increasing the damping ratio

ζ

always lowers the SRS at every frequency.

Mostly true near resonance peaks, where damping cuts amplification; but on the low-frequency rising branch and the high-frequency flat branch the SRS is nearly insensitive to

ζ

, so "always" is too strong.

A "1000g SRS at 500 Hz" number is meaningless without also stating the damping ratio.

True. The SRS value at a frequency changes with

ζ

; comparing a component's capability to an environment requires both be quoted at the same

ζ

(commonly 5%, i.e. Q

= 10

The primary (positive) SRS and residual (negative) SRS can differ.

True. Response during the pulse (primary) and after it ends (residual, free decay) can peak at different values; standards distinguish maximax, primary, and residual SRS for exactly this reason.

Spot the error

"The SRS at

f_{n}

equals the acceleration the base reaches when it vibrates at

f_{n}

The base doesn't "vibrate at

f_{n}

" — a single transient pulse contains all frequencies at once. The SRS at

f_{n}

is the peak response of an oscillator tuned to $f_{n}$ , not a property of the base at that frequency.

"Since

\overset{z}{¨}

is the acceleration we compute, the SRS should plot

max ∣ \overset{z}{¨} ∣

\overset{z}{¨}

is relative acceleration measured from the moving base. What breaks hardware is absolute acceleration

a_{abs} = - 2 ζ ω_{n} \overset{z}{˙} - ω_{n}^{2} z

; plotting

max ∣ \overset{z}{¨} ∣

would give the wrong (base-frame) damage measure.

"The equation of motion is

m \overset{z}{¨} + c \overset{z}{˙} + k z = + ma (t)

The sign is wrong. In the base-relative frame the inertial forcing term is

- ma (t)

— the base acceleration acts like a force opposite to the base motion (the "pushed backward in an accelerating car" effect; see the base-excitation figure above).

"A shorter shock pulse is always less severe because it delivers less energy."

A shorter pulse pushes the knee to higher frequency (

f_{knee} \approx 1/ (π τ)

), so it can be more damaging to high-frequency components (relays, PCBs) even with less total energy — pyroshocks are exactly this: brief but high-frequency-rich.

"To find the SRS I take the FFT of

a (t)

and read off the peaks."

No FFT is involved. You solve the SDOF equation (via Duhamel convolution or numerical integration) for each

f_{n}

, then take the time-domain peak of

∣ a_{abs} (t) ∣

ω_{d} = ω_{n} 1 + ζ^{2}

The sign under the root is wrong: the damped natural frequency is

ω_{d} = ω_{n} 1 - ζ^{2}

, which is slower than the undamped one — damping drags the ringing frequency down, not up.

"The peak amplification of an SDOF at resonance is always about

2 \times

For steady-state sinusoidal drive the peak amplification is

\approx Q = 1/ (2 ζ)

(e.g.

10

at 5% damping). A short transient only partially rings up, so it may show far less than

Q

— the amplification depends on how many cycles the pulse excites.

Why questions

Why do we add

a (t)

\overset{z}{¨}

to get absolute acceleration, and where does the

\overset{z}{¨}

term go?

By definition

a_{abs} = a (t) + \overset{z}{¨}

. Solve the EOM

\overset{z}{¨} + 2 ζ ω_{n} \overset{z}{˙} + ω_{n}^{2} z = - a (t)

for

\overset{z}{¨} = - a (t) - 2 ζ ω_{n} \overset{z}{˙} - ω_{n}^{2} z

and substitute: the two

a (t)

terms cancel exactly, leaving

a_{abs} = - 2 ζ ω_{n} \overset{z}{˙} - ω_{n}^{2} z

— the relative-acceleration term is replaced by the spring and damper terms.

Why does the low-frequency branch of a pulse SRS rise as

f_{n}^{2}

(+40 dB/decade)?

A soft, slow oscillator barely moves during the brief pulse — it behaves like a static spring displaced by the net impulse. Its peak absolute acceleration

ω_{n}^{2} z

scales as

ω_{n}^{2} \propto f_{n}^{2}

, hence the steep rise.

Why do we model real components as SDOF oscillators at all?

Near its fundamental resonance almost any structure behaves like one mass on one spring with one damper (see Modal analysis); the SRS precomputes worst-case response for every possible fundamental frequency so we needn't test each part individually.

Why is the SRS more useful than the raw shock time history for qualification?

The time history is specific to one measurement point and one waveform. The SRS converts it into a frequency-indexed damage measure that lets you directly compare an environment against a component's rated capability at matching

ζ

Why does the knee frequency scale like

1/ (π τ)

for a half-sine of duration

τ

The knee marks where an oscillator's period roughly matches the pulse duration — around one cycle fits inside

τ

. That happens near

f_{n} τ \sim 1/ π

, so shorter pulses move the knee higher.

Why does damping barely change the SRS far below and far above the knee?

On both branches the oscillator either quasi-statically follows the impulse (low

f

) or rigidly tracks the base peak (high

f

); resonant ring-up — the only thing damping strongly suppresses — happens mainly near the knee.

Edge cases

What does the SRS look like as

f_{n} \to 0

(an infinitely soft mount)?

It tends toward zero: an extremely soft oscillator is displaced by the impulse but has vanishing

ω_{n}^{2}

, so its absolute acceleration

\to 0

— this is the ideal of perfect shock isolation.

What happens to the response formula at exactly

ζ = 1

(critical damping)?

The underdamped Duhamel form breaks down because

ω_{d} = ω_{n} 1 - ζ^{2} = 0

; you must use the critically damped solution with a

t e^{- ω_{n} t}

term instead of the sine. The SRS itself stays finite and smooth.

What about the over-damped regime

ζ > 1

— does the SRS still exist?

Yes. Now

1 - ζ^{2} < 0

, so

ω_{d}

becomes imaginary and the sine turns into a sum of real decaying exponentials

e^{(- ζ ω_{n} \pm ω_{n} ζ^{2} - 1) t}

— no oscillation at all, just a sluggish return. The SRS is still

max_{t} ∣ a_{abs} ∣

and is even smoother and lower than the critically damped case, since heavy damping suppresses ring-up entirely; it smoothly approaches the input-peak envelope with no resonant hump.

Can the SRS value ever fall below the input peak acceleration?

Yes — on the low-frequency branch a soft oscillator responds with less than the base peak (de-amplification). Amplification above the input peak only appears near and above the knee.

Is it valid to compare a supplier's 1500g SRS at

ζ = 0.03

to a flight spec of 1800g at

ζ = 0.05

No — different damping means different curves. You must re-derive both at a common

ζ

; a lower-damping rating is generally more conservative (higher for the same pulse), so mixing them can hide or fake margin.

For a very long pulse (

f_{n} τ ≫ 1

) relative to the oscillator, what response does the SDOF give?

The oscillator has time to reach quasi-steady following of the slowly varying base, so its absolute acceleration tracks the input closely — the SRS flattens toward the input peak, the high-frequency asymptote.

What if two shock events overlap in time — does the SRS simply add?

No. The SRS is a nonlinear "max of a response" operation, not a linear transform; you must superpose the time histories first (the SDOF response is linear) and then compute a single SRS of the combined signal.