2.4 Tebranishlar — Masalalar

Jami: 30 ta | Yechim bilan: ✅

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Asosiy Masalalar (1-10)

Masala 1 ⭐⭐

Prujina (k = 200 N/m) ga 0.5 kg massa ulangan. Tebranish davrini toping.

Yechim

$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.5}{200}} = 2\pi\sqrt{0.0025} = 2\pi(0.05) = 0.314$ s

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Masala 2 ⭐⭐

Oddiy mayatnik uzunligi 1 m. Tebranish davri? (g = 10 m/s²)

Yechim

$T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1}{10}} = 2\pi(0.316) = 1.99$ s ≈ 2 s

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Masala 3 ⭐⭐

Tebranish: $x(t) = 0.1\cos(5t)$ m. Amplituda, chastota va davr?

Yechim

$A = 0.1$ m

$\omega = 5$ rad/s

$f = \frac{\omega}{2\pi} = \frac{5}{2\pi} = 0.796$ Hz

$T = \frac{1}{f} = 1.26$ s

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Masala 4 ⭐⭐

Prujina-massa tizimida A = 0.05 m, ω = 10 rad/s, m = 0.2 kg. Maksimal tezlik va tezlanish?

Yechim

$v_{max} = A\omega = 0.05 \times 10 = 0.5$ m/s

$a_{max} = A\omega^2 = 0.05 \times 100 = 5$ m/s²

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Masala 5 ⭐⭐

Tebranuvchi jismning umumiy energiyasi E = 2 J, amplituda A = 0.1 m. Prujina konstantasi?

Yechim

$E = \frac{1}{2}kA^2$

$k = \frac{2E}{A^2} = \frac{2(2)}{0.01} = 400$ N/m

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Masala 6 ⭐⭐

Mayatnik Oyda (g = 1.62 m/s²) tebranmoqda. Yerda T = 2 s bo'lsa, Oydagi davr?

Yechim

$T \propto \sqrt{\frac{1}{g}}$

$\frac{T_{moon}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{moon}}} = \sqrt{\frac{10}{1.62}} = 2.48$

$T_{moon} = 2 \times 2.48 = 4.96$ s

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Masala 7 ⭐⭐

x = 0.04 m da tezlik v = 0.3 m/s. Amplituda A = 0.05 m bo'lsa, burchak chastota?

Yechim

Energiya saqlanishi: $\frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

$\omega^2(A^2 - x^2) = v^2$

$\omega = \frac{v}{\sqrt{A^2 - x^2}} = \frac{0.3}{\sqrt{0.0025 - 0.0016}} = \frac{0.3}{0.03} = 10$ rad/s

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Masala 8 ⭐⭐

Prujinaga 2 kg massa ulangan, f = 5 Hz. Prujina konstantasi?

Yechim

$\omega = 2\pi f = 31.42$ rad/s

$k = m\omega^2 = 2 \times 987.5 = 1975$ N/m ≈ 2000 N/m

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Masala 9 ⭐⭐

Tebranish: $x(t) = 0.08\cos(4t + \pi/3)$ m. t = 0 da pozitsiya va tezlik?

Yechim

$x(0) = 0.08\cos(\pi/3) = 0.08 \times 0.5 = 0.04$ m

$v(t) = -0.08 \times 4 \sin(4t + \pi/3) = -0.32\sin(4t + \pi/3)$

$v(0) = -0.32\sin(\pi/3) = -0.32 \times 0.866 = -0.277$ m/s

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Masala 10 ⭐⭐

Soat mayatniki 1 s davr bilan tebranadi. Uzunligini toping.

Yechim

$T = 2\pi\sqrt{\frac{L}{g}}$

$L = \frac{gT^2}{4\pi^2} = \frac{10 \times 1}{39.48} = 0.253$ m ≈ 25 cm

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O'rtacha Masalalar (11-22)

Masala 11 ⭐⭐⭐

So'nuvchi tebranish: $\zeta = 0.1$, $\omega_0 = 10$ rad/s. Damped chastota va 5 ta tebranishda amplituda qancha kamayadi?

Yechim

$\omega_d = \omega_0\sqrt{1-\zeta^2} = 10\sqrt{1-0.01} = 10 \times 0.995 = 9.95$ rad/s

5 ta tebranish davri: $T_d = \frac{2\pi}{\omega_d} = 0.631$ s

Umumiy vaqt: $t = 5 \times 0.631 = 3.16$ s

Amplituda kamayishi: $\frac{A(t)}{A_0} = e^{-\zeta\omega_0 t} = e^{-0.1 \times 10 \times 3.16} = e^{-3.16} = 0.042$

Amplituda 95.8% kamaydi (4.2% qoldi).

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Masala 12 ⭐⭐⭐

Critically damped tizim: m = 1 kg, k = 100 N/m. Damping koeffitsienti c?

Yechim

$\omega_0 = \sqrt{\frac{k}{m}} = 10$ rad/s

Critical damping: $\zeta = 1$

$c = 2\zeta\sqrt{km} = 2 \times 1 \times \sqrt{100 \times 1} = 20$ N·s/m

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Masala 13 ⭐⭐⭐

Rezonans: $\omega_0 = 50$ rad/s, $\zeta = 0.05$. Q-factor va bandwidth?

Yechim

$Q = \frac{1}{2\zeta} = \frac{1}{0.1} = 10$

Bandwidth: $\Delta\omega = \frac{\omega_0}{Q} = \frac{50}{10} = 5$ rad/s

Yoki: $\Delta f = \frac{f_0}{Q} = \frac{50/(2\pi)}{10} = 0.796$ Hz

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Masala 14 ⭐⭐⭐

Majburiy tebranish: $\omega_0 = 20$ rad/s, $\zeta = 0.1$, $F_0 = 10$ N, k = 400 N/m. Haydash chastotasi $\omega = 18$ rad/s da amplituda?

Yechim

$r = \frac{\omega}{\omega_0} = \frac{18}{20} = 0.9$

$X = \frac{F_0/k}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}$

$= \frac{10/400}{\sqrt{(1-0.81)^2 + (0.18)^2}}$

$= \frac{0.025}{\sqrt{0.0361 + 0.0324}}$

$= \frac{0.025}{\sqrt{0.0685}} = \frac{0.025}{0.262} = 0.095$ m

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Masala 15 ⭐⭐⭐

Fizik mayatnik: massa 2 kg, massa markazigacha masofa 0.3 m, inertsiya momenti 0.2 kg·m². Davr?

Yechim

$T = 2\pi\sqrt{\frac{I}{mgh}} = 2\pi\sqrt{\frac{0.2}{2 \times 10 \times 0.3}} = 2\pi\sqrt{\frac{0.2}{6}} = 2\pi(0.183) = 1.15$ s

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Masala 16 ⭐⭐⭐

Avtomobil suspension: m = 300 kg (bir g'ildirak uchun), k = 30000 N/m, c = 3000 N·s/m. Damping nisbati va xarakteristika?

Yechim

$\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{100} = 10$ rad/s

$\zeta = \frac{c}{2\sqrt{km}} = \frac{3000}{2\sqrt{30000 \times 300}} = \frac{3000}{6000} = 0.5$

$\zeta = 0.5 < 1$ → Underdamped

Bu yaxshi — tez javob beradi, lekin juda ham tebranmaydi.

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Masala 17 ⭐⭐⭐

Dron vibratsiya: propeller 6000 rpm. Chastota Hz da va agar frame tabiiy chastotasi 90 Hz bo'lsa, rezonans bormi?

Yechim

$f = \frac{6000}{60} = 100$ Hz

Frame: 90 Hz

Farq: 10 Hz — rezonans yo'q, lekin yaqin!

Xavfli zona: $90 \pm \frac{90}{Q}$ — Q katta bo'lsa tor banda.

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Masala 18 ⭐⭐⭐

Prujina ketma-ket ulangan: $k_1 = 100$ N/m, $k_2 = 200$ N/m. Ekvivalent k va 0.5 kg massa bilan davr?

Yechim

Ketma-ket: $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200}$

$k_{eq} = \frac{200}{3} = 66.7$ N/m

$T = 2\pi\sqrt{\frac{0.5}{66.7}} = 2\pi(0.0866) = 0.544$ s

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Masala 19 ⭐⭐⭐

Prujina parallel ulangan: $k_1 = 100$ N/m, $k_2 = 200$ N/m. Ekvivalent k?

Yechim

Parallel: $k_{eq} = k_1 + k_2 = 100 + 200 = 300$ N/m

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Masala 20 ⭐⭐⭐

So'nuvchi tebranishda energiya 10 tebranishda 50% kamaydi. Damping nisbati?

Yechim

Energiya: $E \propto A^2$

$\frac{E(t)}{E_0} = e^{-2\zeta\omega_0 t} = 0.5$

$-2\zeta\omega_0 t = \ln(0.5) = -0.693$

10 tebranish: $t = 10T = \frac{20\pi}{\omega_0}$

$2\zeta\omega_0 \times \frac{20\pi}{\omega_0} = 0.693$

$40\pi\zeta = 0.693$

$\zeta = \frac{0.693}{125.66} = 0.0055$

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Masala 21 ⭐⭐⭐

Beat fenomeni: $f_1 = 440$ Hz, $f_2 = 442$ Hz. Beat chastotasi?

Yechim

$f_{beat} = |f_1 - f_2| = |440 - 442| = 2$ Hz

Sekundiga 2 marta kuchayadi-so'nadi.

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Masala 22 ⭐⭐⭐

Robot qo'li end effector tebranmoqda: A = 1 mm, f = 50 Hz. Maksimal tezlanish (g da)?

Yechim

$\omega = 2\pi f = 314$ rad/s

$a_{max} = A\omega^2 = 0.001 \times 98596 = 98.6$ m/s²

$\frac{a_{max}}{g} = \frac{98.6}{10} = 9.86$ g

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Murakkab Masalalar (23-30)

Masala 23 ⭐⭐⭐⭐

PID controller overdamped javob berdi. $\zeta = 1.5$, $\omega_0 = 10$ rad/s. Settling time (2% ga)?

Yechim

Overdamped uchun: $s_{1,2} = -\omega_0(\zeta \pm \sqrt{\zeta^2-1}) = -10(1.5 \pm 1.118)$

$s_1 = -26.18$, $s_2 = -3.82$

Dominant pole: $s_2 = -3.82$

Settling time: $t_s \approx \frac{4}{|s_2|} = \frac{4}{3.82} = 1.05$ s

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Masala 24 ⭐⭐⭐⭐

Dron IMU vibratsiya filtri: cutoff 20 Hz, tabiiy chastota 100 Hz. 100 Hz da signal qancha kamayadi (2nd order)?

Yechim

2nd order low-pass: $|H(j\omega)| = \frac{1}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}$

$\zeta = 0.707$ (Butterworth) qabul qilsak:

$r = \frac{100}{20} = 5$

$|H| = \frac{1}{\sqrt{(1-25)^2 + (1.414 \times 5)^2}} = \frac{1}{\sqrt{576 + 50}} = \frac{1}{25} = 0.04$

$20\log(0.04) = -28$ dB

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Masala 25 ⭐⭐⭐⭐

Ikki bog'langan mayatnik: $m_1 = m_2 = 1$ kg, $L_1 = L_2 = 1$ m, orasida prujina k = 10 N/m. Normal modlar chastotalari?

Yechim

Harakat tenglamalari: $m\ddot{x}_1 = -\frac{mg}{L}x_1 - k(x_1 - x_2)$ $m\ddot{x}_2 = -\frac{mg}{L}x_2 - k(x_2 - x_1)$

$\omega_g^2 = \frac{g}{L} = 10$

Mode 1 (in-phase): $\omega_1 = \sqrt{\omega_g^2} = \sqrt{10} = 3.16$ rad/s

Mode 2 (out-of-phase): $\omega_2 = \sqrt{\omega_g^2 + \frac{2k}{m}} = \sqrt{10 + 20} = \sqrt{30} = 5.48$ rad/s

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Masala 26 ⭐⭐⭐⭐

Seismometr: m = 0.1 kg, k = 40 N/m, c = 0.4 N·s/m. 1 Hz va 10 Hz signallarni qanday qayd qiladi?

Yechim

$\omega_0 = \sqrt{400} = 20$ rad/s → $f_0 = 3.18$ Hz

$\zeta = \frac{0.4}{2\sqrt{40 \times 0.1}} = \frac{0.4}{4} = 0.1$

1 Hz ($\omega = 6.28$): $r = 0.314$ $|H| = \frac{r^2}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}} = \frac{0.099}{\sqrt{0.81 + 0.004}} = 0.11$

10 Hz ($\omega = 62.8$): $r = 3.14$ $|H| = \frac{9.87}{\sqrt{81.5 + 3.94}} = \frac{9.87}{9.24} = 1.07$

Yuqori chastotalar yaxshi, past chastotalar yomon.

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Masala 27 ⭐⭐⭐⭐

Torsion pendulum: I = 0.05 kg·m², torsion constant κ = 2 N·m/rad. 10° dan qo'yib yuborildi. t = 1 s da burchak?

Yechim

$\omega_0 = \sqrt{\frac{\kappa}{I}} = \sqrt{\frac{2}{0.05}} = \sqrt{40} = 6.32$ rad/s

$\theta(t) = \theta_0 \cos(\omega_0 t) = 10° \cos(6.32 \times 1) = 10° \cos(6.32)$

$\cos(6.32) = 0.998$ (6.32 rad ≈ 362°)

$\theta(1) = 10° \times 0.998 = 9.98°$

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Masala 28 ⭐⭐⭐⭐

Quadcopter altitude controller: $\omega_n = 5$ rad/s, $\zeta = 0.7$ maqsad. Step response overshoot va settling time?

Yechim

Overshoot: $M_p = e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^2}}} = e^{-\frac{3.14 \times 0.7}{0.714}} = e^{-3.08} = 0.046 = 4.6\%$

Settling time (2%): $t_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.7 \times 5} = 1.14$ s

Rise time: $t_r \approx \frac{1.8}{\omega_n} = 0.36$ s

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Masala 29 ⭐⭐⭐⭐

Parametrik rezonans: mayatnik support vertikal $y(t) = Y\cos(2\omega_0 t)$ bilan tebranmoqda. Y qancha bo'lganda beqaror?

Yechim

Mathieu tenglamasi — parametrik rezonans sharti:

Beqarorlik $2\omega_0$ da boshlanadi, agar:

$\frac{Y\omega_0^2}{g} > \frac{4\zeta}{\omega_0}$ (taxminan)

Damping past bo'lganda ($\zeta \to 0$), juda kichik Y ham beqarorlik keltirib chiqaradi.

Kritik: $Y > \frac{4\zeta g}{\omega_0^3}$

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Masala 30 ⭐⭐⭐⭐

Half-power bandwidth: rezonans $f_0 = 100$ Hz, Q = 50. Bandwidth va -3dB nuqtalar?

Yechim

$\Delta f = \frac{f_0}{Q} = \frac{100}{50} = 2$ Hz

-3dB nuqtalar: $f_1 = f_0 - \frac{\Delta f}{2} = 100 - 1 = 99$ Hz $f_2 = f_0 + \frac{\Delta f}{2} = 100 + 1 = 101$ Hz

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✅ Tekshirish Ro'yxati

• 1-10: Asosiy SHM
• 11-22: Damping va rezonans
• 23-30: Murakkab tizimlar

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Keyingi Qadam

📖 2.5 Qattiq Jism Mexanikasi