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


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Fizika 2 qisim

Javob: D) 5.
27.5. Moddiy nugtaning tebranish gonuni π‘₯ = 0.03 sin

2πœ‹π‘‘
ko’rinishda berilgan. Uning T = 3T/4
𝑇
paytdagi siljishi modulini toping (m).
A) 0.05 B) 0.04 C) 0,03 D) 0.02
Berilgan:
2πœ‹π‘‘
π‘₯ = 0.03 sin
𝑇

Ξ€ =
|π‘₯| =?


Yechilishi: Berilgan garmonik tenglamada vagtning o'rniga berilgan givmatini go’vamiz:

2πœ‹π‘‘
|π‘₯| = 0.03 sin = 0.03 sin
𝑇
Javob: C) 0.03.
2πœ‹ 3𝑇
β‹… = |βˆ’0.03| = 0.03β€ˆπ‘š
𝑇 4

27.6. Nugtaning harakar tenglamnasi π‘₯ = 0.3 sin 0.1πœ‹π‘‘, ko'rinishda yoziladi.Harakatning qandayligi va uni tavsifiovchi kattaliklarni aniqlang.

  1. radiusi 0,3 bo’lgan aylana bo'ylab tekis harakat, chizigli tezligi 0.1m/s.

  2. Tebranma harakat, amplitudasi 3 sm, chastotasi 0,1 Hz.

C tebranma harakat, amplitudasi 30 sm, davri 20 s.
D) radiusi 30 sm bo'lgan aylana bo'ylab tekis harakat, aylanish davri
2,0 s.
Yechilishi: Berilgan harakat tenglamasiga asosan:
π‘₯π‘š = 0.3π‘š = 30π‘ π‘š; πœ” = 0.1πœ‹.
Tebranish davrini siklik chastotaga bog’lanish fodasidan t topamiz:
2πœ‹ 2πœ‹

Ξ€ = = = 20 𝑠 πœ” 0.1πœ‹
Javob: C) tebranma harakat, amplitudasi 30 sm, davri 20 s.
27.7. Amplitudasi 20 sm, tebranish davri 5 s, boshlang’ich fazasi 0 ga teng bo'lgan garmonik tebranish tenglamasini yozing,

A)π‘₯ = 0.2 sin 0. 4πœ‹π‘‘


B)π‘₯ = 0.2 sin 5 πœ‹π‘‘

C) π‘₯ = 20 sin 0. 2πœ‹π‘‘


πœ‹
D)π‘₯ = 20 sin (0.4πœ‹π‘‘ + )
4
Yechilishi: Berilgan kattaliklarni garmonik harakat tenglamasiga





qo'yamiz:

π‘₯ = π‘₯π‘š sin(πœ”π‘‘ + πœ‘0) ;
2πœ‹ 2πœ‹
πœ” = = = 0.4πœ‹
𝑇 5
O'miga go'ysak, quyidagini olamiz.
π‘₯ = 0.2 sin( ) 0.4πœ‹π‘‘

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