Berilgan: K=125 N/m t=8s N=5 m=? Yechilishi

Javob:A)𝑇_𝑠
28.7. Oyda erkin tushish tezlanishi yerdagiga nisbatan 6 marta kichik.Oyda matematik mayatnikning tebranish davri yerdagi tebranish davridan qanday farq qiladi?
A) 6 marta katta B) marta katta C) marta kichik D) 6 marta kichik
Berilgan:
𝑔_𝑦𝑒𝑟 = 6𝑔_𝑜𝑦 Yechilishi: Matematik mayatnikning tebranish davrini topish
𝑇𝑦𝑒𝑟
=? formulasidan foydalanamiz:
𝑇𝑜𝑦
𝑙
𝑇𝑦𝑒𝑟 = 2𝜋√𝑔𝑦𝑒𝑟, 𝑇𝑜𝑦 = 2𝜋√𝑔𝑜𝑦,
Tengalamani nisbatini olamiz.
𝑇 2𝜋 𝑦𝑒𝑟 √ , 𝑇𝑜𝑦 𝑇𝑦𝑒𝑟
𝑔𝑜𝑦
Javob: B) marta katta.
28.8. Agar Y erda tebranish chastotasi 0,5 Hz bo'Igan matematik mayatnik Oyga olib chiqilsa, u qanday chastotada tebranadi? Oyda erkin tushish tezlanishi Yerdagidan 6 marta kichik.
A) 0,2 B) 0,3 C) 0.6 D) 0.8 Berilgan:

𝑔𝑦𝑒𝑟 = 6𝑔𝑜𝑦	Yechilishi: Matematik mayatnikning tebranish chastotasini topish
𝜈𝑦𝑒𝑟 = 0.5𝐻𝑧	formulasidan foydalanamiz:

𝜈𝑜𝑦 =? 𝜈𝑦𝑒𝑟 , 𝜈𝑜𝑦
2𝜋2𝜋
Tenglamalarninisbatini olamiz
𝜈𝑦𝑒𝑟

√
𝜈𝑜𝑦 ^𝑔𝑜𝑦
𝑙
2𝜋√𝑔𝑜𝑦
𝜈_𝑦𝑒𝑟 0.5
𝜈_𝑜𝑦 𝐻𝑧
Javob:A)0.2
28.9. Mayatnikli soat Yerdan Oyga ko'chirilsa, uning yurishi ganday o'zgaradi? Oyda erkin tushish tezlanishi 1,6 m/s ga teng

1. 
   6 marta sekinlashadi
   
2. 
   6 marta tezlashadi
   
3. 
   2,5 marta tezlashadi D) 2,5 marta sekinlashadi
   

Berilgan:
𝑔_𝑦𝑒𝑟 = 6𝑔_𝑜𝑦 Yechilishi: Matematik mayatnikning tebranish dav ini topish 𝑇_𝑦𝑒𝑟 = 𝑇_𝑜𝑦 formulasidan foydalanamiz:

𝑙𝑦𝑒𝑟𝑙
𝑙𝑜𝑦 =? 𝑇_𝑦𝑒𝑟 = 2𝜋 , 𝑇_𝑜𝑦 = 2𝜋√𝑔𝑜𝑦, Tenglamalar nisbatini olamiz:

𝑇𝑦𝑒𝑟
;
𝑇𝑜𝑦
1
𝑇𝑜𝑦 = 𝑇𝑦𝑒𝑟 = 2.5𝑇𝑦𝑒𝑟
0.4
Javob: D) 2.5 marta sekinlashadi.
28.10. Uzunligi 16 sm bo'lgan mayatnik biror vagt ichida S marta tebrandi. Ikkinchi mayatnik shu vaqt ichida 10 marta tebransa, uning uzunligi qgnday (sm)?

1. 
   24
   
2. 
   16
   
3. 
   12
   
4. 
   4
   

Berilgan:
𝑡₁ = 𝑡₂ = 𝑡 Yechilishi: Matematik mayatnikning tebranish davrini topish formulasidan
𝑁₁ = 5 foydalanamiz:
1
𝑁₂ = 10 𝑇₁ 𝑁 ₂,

Tenglamalar nisbatini olamiz	𝑙1 = 16 𝑠𝑚 𝑙2 =?
𝑇	𝑡 2𝜋 𝑔 𝑙 𝑁 𝑁

₁ √𝑙12 𝑡1 𝑁 12
𝑇₂ 2𝜋√𝑙𝑔2 𝑁₂
tenglamani kvadratga oshiramiz:
𝑙₁ 𝑙₁ 16𝑠𝑚
𝐶2 = 4 𝑙2 = 4 = 4 = 4𝑠𝑚.
Javob: D) 4.
28.11. Matematik mayatnikning tebranish qonuni𝑥 = 0.1 sin 5 𝑡(𝑚) ko'rinishga ega.
Mayatnikning uzunligi necha metr?𝑔 = 10𝑚/𝑠²

1. 
   5
   
2. 
   2,5
   
3. 
   0,5
   
4. 
   0,4
   

Berilgan:
𝑥 = 0.1 sin 5 𝑡(𝑚) Yechilishi: Tebranish gonuni 𝑥 = 𝑥_𝑚𝑠𝑖𝑛𝜔𝑡 bilan tenglamani 𝑔 =
10𝑚/𝑠² solishtiramiz, = 5 rad/s.
l=? Matematik mayatnikning tebranish davrini topishformulasidan foydalanamiz.
1 𝑔
𝑇 𝜔 ₂ ; 𝑙 ⁼ _𝜔 2 = = 0.4 𝑚.
Javob: D) 04
28.12. Matematik mayatnik uzunligi 8 sm ortganda, tebranishlar davri 0,1 s oshdi.
Tebranishlaring dastlabki davri qanday. ( 𝜋² = 10.)
A) 1,4 C)1,7 B) 1.6 D)1,55
Berilgan:
∆𝑙 = 8𝑠𝑚 Yechilishi: Matematik mayatnikning tebranish davrini topish formulasidan
∆𝑇 = 0.1𝑠 foydalanamiz:
𝑇₁ =?
𝑇₁

Uzunlikni topamiz:𝑙₁ = 𝑔_4𝜋𝑇122 ; 𝑙2 = 𝑔4𝜋𝑇222 ;
𝑙₂ = 𝑙₁ + ∆𝑙; 𝑇₂ = 𝑇₁ + ∆𝑇;
∆𝑙 = 𝑙₂ − 𝑙₁ = 𝑔4𝜋𝑇22₂ 𝑔4𝜋𝑇12 𝑔 2 − 𝑇12);
− 2 = 4𝜋2 (𝑇2
𝑔 𝑔
∆𝑙 = 4 𝜋₂ ((𝑇₁ + ∆𝑇)² − 𝑇₁²) = _{4 𝜋}2 (𝑇₁² + 2𝑇₁∆𝑇 + ∆𝑇² − 𝑇₁²);
𝑔 ²);
∆𝑙 = 4𝜋₂ (2𝑇₁∆𝑇 + ∆𝑇
2𝑇∆𝑇 = 4 𝜋²∆𝑙 − ∆Τ₂ = 4 𝜋²∆𝑙 − 𝑔∆Τ² ;
𝑔 𝑔
55 𝑠
Javob: D) 1,55.
28.13. Massasi 50 g bo'lgan po'lat sharchali mayatnikning tebranish davri 2 s. Sharcha ostiga magnit joylashtirilganda, mayatnikning tebranish davri 1 s gacha kamaydi. Sharchaning magnitga tortilish kuchi qanday?

1. 
   0.5
   
2. 
   1
   
3. 
   5
   
4. 
   1,5
   

Berilgan:

m = 50 g ,nisbat olamiz:
𝑔 𝑔+
Τ₁ = 2𝑠
𝑇¹
Τ₂ = 1𝑠 _𝑇2 == 2

𝑎
F=?
kvadratga oshiramiz:g+a=4 g, a=3g=30m/ 𝑠²
F=ma=0.05*30=1.5N
Javob: D) 1,5.
28.14. Matematik mayatnik 7 sm amplituda va 2 s davr bilan garmonik tebranmoqda. Uning koordinatasining vagt bo’ icha o'zgarish qonunini toping.
A) x = 0.7 sin 𝜋𝑡 B)x=0.07sin 𝜋 𝑡
C) x = 7 sin ttB D) x=7 sin2𝜋𝑡