Exercises — Shock response spectrum (SRS)

Yeh Shock response spectrum (SRS) definition ke direct restatements hain. Key relation $Q=1/(2\zeta )$ deti hai $Q=1/(2\times 0.05)=10$. Dekho Quality factor Q and damping.

Yeh woh peak absolute acceleration hai jo ek single SDOF oscillator — $f_n=2000$ Hz (5% damping ke saath, yaani $\zeta =0.05$) par tuned — experience karega agar tum iske base ko is shock se drive karo. Yeh input peak nahi hai, aur FFT amplitude nahi hai — yeh ek imaginary test oscillator ki response hai. (Yeh parent note ki Mistake 1 ka core hai.)

(a) ${\omega }_n=2\pi f_n=2\pi (250)=1570.8\text{rad/s}$. Humne kya kiya / kyun: ${\omega }_n$ radians per second count karta hai, $f_n$ whole cycles per second count karta hai, aur ek cycle $2\pi$ radians hoti hai — isliye $2\pi$ ka factor lagta hai. (b) $T_n=1/f_n=1/250=0.004\text{s}=4\text{ms}$. Yeh ek full bounce ka waqt hai.

Step 1 — branch check karo. Compute karo $f_n\tau =500\times 0.0002=0.1$. Kyunki $0.1\ll 1$ hai, oscillator pulse ke dauran sirf daswan hissa bounce manage karta hai — hum firmly rising (low-frequency) branch par hain, toh asymptote valid hai. Step 2 — plug in karo. $\text{SRS}\approx A_0\frac{{\pi }^2{f}_n^2{\tau }^2}{2}=300\cdot \frac{{\pi }^2(500{)}^2(0.0002{)}^2}{2}$ $=300\cdot \frac{{\pi }^2\cdot 250000\cdot 4\times 10^{-8}}{2}=300\cdot \frac{{\pi }^2\cdot 0.01}{2}\approx 300\cdot 0.04935\approx 14.8g$ Iska matlab: knee se bahut neeche oscillator barely respond karta hai — 300g input mein se sirf ~15g.

$f_{\text{knee}}\approx \frac{1}{\pi \tau }=\frac{1}{\pi (0.0002)}=\frac{1}{6.283\times 10^{-4}}\approx 1592\text{Hz}.$ Matlab: ~1.6 kHz se neeche tum rising branch par ho; uske upar curve peak ki taraf flatten hoti hai. Upar wale figure par yahi exactly woh jagah hai jahan violet dashed line khadi hai. Pulse ko aadha karna: $\tau \to 0.1\text{ms}$ deta hai $f_{\text{knee}}\approx 1/(\pi \cdot 10^{-4})\approx 3183\text{Hz}$ — knee double ho jaati hai. Ek shorter, sharper bang danger zone ko higher frequency par push karta hai. Isliye Pyroshock environments (bahut short, sharp) high-frequency components jaise PCBs aur relays ko threaten karte hain.

Har ek ke liye branch number compute karo:

• A: $f_n\tau =400\times 0.0002=0.08\Rightarrow$ rising branch par gehrai mein.
• B: $f_n\tau =1600\times 0.0002=0.32\Rightarrow$ bilkul knee par ($f_{\text{knee}}\tau =1/\pi \approx 0.318$). B kyun jeetta hai: rising branch par SRS $f_n^2$ ki tarah badhta hai. B ki frequency A ki 4 guna hai, toh sirf asymptote par B already $4^2=16\times$ zyada hoga; aur B additionally knee par baitha hai jahan oscillator almost ek full cycle ke liye ring up karta hai, real amplification add karta hai. B bahut zyada dekhega.

Figure padho. Figure s01 par, navy dot (A: 400 Hz) climb ke flat bottom-left ke paas baitha hai, jabki violet dot (B: 1600 Hz) upar, bilkul knee par — woh 16×-plus gap ka picture jo humne abhi argue kiya. Figure s02 dikhata hai kyun: iska horizontal axis time ms mein hai aur vertical axis normalized response hai; navy trace (low $f_n$) shaded pulse ke dauran barely wiggle karta hai, jabki magenta trace (near-knee $f_n$) visibly ring up karta hai.

(a) Branch. $f_n\tau =200\times 0.0005=0.1$. Yeh $\ll 1$ hai → rising branch, toh oscillator pulse ke dauran barely move karta hai aur response chota hai (amplified nahi). (Single-degree-of-freedom systems language: pulse khatam ho jaati hai board ke ek swing complete karne se pehle.) (b) Estimate — branch ka apna formula use karo. Kyunki hum rising branch par hain, humein low-frequency asymptote use karni chahiye, koi made-up amplification factor nahi: $\text{SRS}\approx A_0\frac{{\pi }^2{f}_n^2{\tau }^2}{2}=500\cdot \frac{{\pi }^2(200{)}^2(0.0005{)}^2}{2}.$ Pieces compute karo: $(200{)}^2=40000$, $(0.0005{)}^2=2.5\times 10^{-7}$, product $=0.01$. Phir $\text{SRS}\approx 500\cdot \frac{{\pi }^2\cdot 0.01}{2}=500\cdot 0.04935\approx 24.7g.$ Toh board sirf ≈25g dekhta hai, 500g input se bahut neeche — exactly yahi "rising branch" ka matlab hai. (Yeh tempting lekin galat "2×500 = 1000g" ko correct karta hai: rising branch par is gehrai mein 1 se upar ka factor nahi aa sakta.) (c) Worst case. Peak knee ke paas rehti hai: $f_{\text{knee}}\approx 1/(\pi \tau )=1/(\pi \cdot 0.0005)\approx 637\text{Hz}$. Wahan tuned board sabse zyada ring up karega (wahan, aur sirf wahan, amplification $Q$-limited value ki taraf climb karti hai). Engineering action: board ka $f_n$ knee se bahut neeche rakhna (jaise yahan hai) — iske apne dynamics use protect karte hain. Danger tab appear hota hai jab redesign accidentally $f_n$ ko ~637 Hz ki taraf push kare; tab damping add karo (potting compound $\zeta$ badhata hai, peak cap karta hai) ya shock-isolate karo. Dekho Force limiting in shock testing aur Shock testing methods.

(a) Same $\zeta$ (5%), toh hum directly compare karte hain: capability 1500g < requirement 1800g → as-is qualified NAHI. (b) Environment ko rising branch par $\propto f_n^2$ treat karte hue aur 2000 Hz par 1800g pin karte hue: $\text{SRS}(800)\approx 1800g{\left(\frac{800}{2000}\right)}^2=1800g\times (0.4{)}^2=1800g\times 0.16=288g.$ Ab $288g<1500g$ → margin ke saath qualified (margin $=1500/288\approx 5.2\times$). $f_n$ ko knee se lower karne se environment dramatically drop ho gayi. Isliye isolation/soft-mounting go-to fix hai.

Setup — teen symbols yaad karo. $a(t)$ input base acceleration hai, $z(t)$ relative displacement (moving base se measure ki gayi mass position), aur $a_{\text{abs}}$ woh absolute acceleration hai jise hum SRS ke liye peak-detect karte hain. Input half-sine $a(t)=A_0\sin (\pi t/\tau )$ for $0\le t\le \tau$, phir $a=0$. Base-relative displacement obey karta hai (parent EOM se, $\zeta \to 0$ ke saath kyunki rising branch par is gehrai mein damping barely matter karti hai): $\ddot{z}+{\omega }_n^{2}z=-a(t).$

Kyun $a_{\text{abs}}=-{\omega }_n^{2}z$ (justified step). Parent ke exact relation $a_{\text{abs}}=-2\zeta {\omega }_n\dot{z}-{\omega }_n^{2}z$ se shuru karo. Response peak par dono terms compare karo: free ringing satisfy karta hai $\dot{z}\sim {\omega }_{n}z$, toh damping term hai $\sim 2\zeta {\omega }_n\cdot {\omega }_{n}z=2\zeta {\omega }_n^{2}z$ — stiffness term ka $2\zeta$ fraction. $\zeta =0.03$–$0.05$ ke liye yeh sirf 6%–10% hai, estimate ke liye negligible. Ise drop karne par milta hai $a_{\text{abs}}\approx -{\omega }_n^{2}z$: absolute acceleration spring ke stiffness term se set hoti hai. (Physically: lightly damped oscillator ke liye, mass par mehsoos hone wali almost sari force spring ke through aati hai, aur spring force $=kz=m{\omega }_n^{2}z$, toh acceleration $={\omega }_n^{2}z$.)

Step 1 — kyun pulse ek impulse ki tarah act karti hai (WHAT/WHY). Jab $f_n\tau \ll 1$ ho, poori pulse par ${\omega }_{n}t\le {\omega }_n\tau =2\pi f_n\tau \ll 1$, toh $z$ aur iska restoring term ${\omega }_n^{2}z$ pulse ke dauran negligible hain; equation essentially $\ddot{z}\approx -a(t)$ hai. Ek baar integrate karne par, pulse system ko ek velocity change deti hai $\dot{z}(\tau )=-{\int }_0^{\tau }adt=-\Delta v,\Delta v={\int }_0^{\tau }{A}_0\sin \left(\frac{\pi t}{\tau }\right)dt=A_0\frac{2\tau }{\pi },$ jabki $z(\tau )\approx 0$ chodti hai (displacement ek tiny quantity ka ek aur integration hai). Kaisa dikhta hai: oscillator still khada hai, phir kick milti hai — figure s02 mein, near-knee trace shaded pulse ke theek baad ring karna shuru karta hai.

Step 2 — kick ke baad free ringing. $t>\tau$ ke liye system initial conditions $z\approx 0$, $\dot{z}\approx -\Delta v$ se freely ring karta hai: $z(t)=-\frac{\Delta v}{{\omega }_n}\sin ({\omega }_n(t-\tau )).$ Iska amplitude hai $z_{\max }=\Delta v/{\omega }_n$.

Step 3 — impulse-level estimate. Kyunki $a_{\text{abs}}=-{\omega }_n^{2}z$ hai, peak hai ${\text{SRS}}_{\text{impulse}}={\omega }_n^2{z}_{\max }={\omega }_n^2\frac{\Delta v}{{\omega }_n}={\omega }_n\Delta v=2\pi f_n\cdot A_0\frac{2\tau }{\pi }=4A_0{f}_n\tau .$ Yeh already prove karta hai ki SRS $f_n$ ke proportional hai aur $f_n\to 0$ par vanish karta hai. Lekin pure impulse woh small correction throw away karta hai ki $z$ exactly zero nahi hai $t=\tau$ par; exact half-sine ise rakhti hai, aur woh correction leading behaviour ko $\propto f_n$ se $\propto f_n^2$ mein badal deta hai.

Step 4 — exact residual solution (missing integration). $\ddot{z}+{\omega }_n^{2}z=-A_0\sin (\pi t/\tau )$ solve karo exactly $0\le t\le \tau$ par $z(0)=\dot{z}(0)=0$ ke saath. $\Omega =\pi /\tau$ likhte hue, particular + homogeneous solution hai $z(t)=\frac{-A_0}{{\omega }_n^2-{\Omega }^2}\left[\sin (\Omega t)-\frac{\Omega }{{\omega }_n}\sin ({\omega }_{n}t)\right].$ Pulse ke baad ($t>\tau$) motion free ringing hai jiski amplitude $z(\tau ),\dot{z}(\tau )$ se fix hoti hai. ${\omega }_n\ll \Omega$ ke liye (yaani $f_n\tau \ll 1$) expand karo $\frac{1}{{\omega }_n^2-{\Omega }^2}\approx -\frac{1}{{\Omega }^2}=-\frac{{\tau }^2}{{\pi }^2}$. Is limit mein post-pulse ring amplitude evaluate karne par milta hai $z_{\max }\approx \frac{A_0}{{\Omega }^2}\cdot \frac{{\pi }^2{f}_n^2{\tau }^2}{2{\omega }_n^2}\cdot {\omega }_n^2\cdot \frac{1}{{\omega }_n^2}⟹{\omega }_n^2{z}_{\max }\approx A_0\frac{{\pi }^2{f}_n^2{\tau }^2}{2}.$ Cleanly: residual ring amplitude ek leading factor ${\omega }_n^2$ carry karta hai ($\sin (\Omega t)$ term se pulse end par evaluate kiya gaya), aur $a_{\text{abs}}=-{\omega }_n^{2}z$ mein stiffness factor ${\omega }_n^2$ se multiply karke, phir ${\omega }_n^2=4{\pi }^2{f}_n^2$ insert karte hue, sab kuch collapse ho jaata hai $\boxed{\text{SRS}={\omega }_n^2{z}_{\max }⟶A_0\frac{{\pi }^2{f}_n^2{\tau }^2}{2}}$ Yeh $\propto f_n^2$ hai (crude impulse $\propto f_n$ se steeper) — true low-frequency behaviour.

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

(a) Kyun flatten hoti hai (WHAT/WHY). Jab $f_n\tau \gg 1$ ho, oscillator ka period $T_n=1/f_n$ pulse se bahut chota hota hai. Base ke slow, gentle rise aur fall ke dauran, aisa stiff, fast oscillator har instant par base ke saath quasi-static equilibrium mein settle hone ka time rakhta hai: mass simply base ko follow karta hai. Phir relative displacement $z$ chota hota hai aur component ki absolute acceleration base acceleration ke barabar hoti hai. Iska peak isliye input peak hai: $\text{SRS}\to A_0(f_n\tau \gg 1).$ Input peak par ek horizontal line — figure s01 par far right ka flat magenta plateau. Yeh essential high-frequency edge case hai: SRS hamesha ke liye nahi badhti reh sakti; yeh $A_0$ par saturate hoti hai (knee ke theek baad settle hone se pehle ek small overshoot bump hoti hai).

(b) $Q$-limited peak (alag scenario). Agar base ko exactly $f_n$ par sinusoidally aur hamesha ke liye drive kiya jaaye (steady-state resonance, transient pulse nahi), toh response kaafi cycles mein build up hokar amplification tak pohonchega $\frac{a_{\text{abs}}}{A_0}{|}_{\text{resonance}}=Q=\frac{1}{2\zeta }.$ $\zeta =0.05$ ke liye yeh hai $Q=1/(2\times 0.05)=10$ — 10× amplification. Contrast: ek short shock pulse drive ko full $Q$ tak pahunchne ke liye kabhi itni der sustain nahi karta; half-sine plateau $A_0$ ke paas saturate hoti hai (amplification ≈1, knee ke theek baad ek modest overshoot ≈1.7× ke saath), 10× nahi. Full $Q$ sirf long, resonant input ke peak ke liye appear hota hai — dekho Quality factor Q and damping. Isliye blindly shock input ko $Q$ se multiply karna plateau over-predict karta hai.

Pehle $f_n\tau$ compute karo ::: agar $\ll 1$ hai tum quadratic (+12 dB/oct) rising branch par ho aur response chota hai; $1/\pi$ ke paas tum knee par ho (max amplification); agar $\gg 1$ hai tum flat plateau par ho jo input peak $A_0$ track karta hai.