Javob: A) o'zgarmaydi.
28.3. Matematik mayatnik uzunligi 4 marta orttirilsa, uning tebranishlar davri qanday o'zgaradi? A) 4 marta kamayadi B) 2 marta ortadi C) 2 marta kamayadi D) 4 marta ortadi
Berilgan:
π1 = 4π1 Yechilishi:Matematik mayatnikning tebranish davrini topish formulasidan
π1 f =? oydalanimiz:
π2
π1 = 2π , π 2.
Tenglamalar nisbatini olamiz:
π1
, π2 = 2π1
π2
Javob: B) 2 marta ortadi.
28.4. Matematik mayatnikning uzuntigi 2,5 m, unga osilgan sharchaningmassi 100 g.Tebranish davri qandey?
A) 3.14 B) 6,28 C) 62.8 D) 31.4
Berilgan:
l=2.5 m Yechilishi: Matematik mayatnikning tebranish davrini topish formulasidan
m = 100 g foydalanamiz:
T=? π π
Javob: A) 3,14.
28.5. Matematik mayatnikning uzunligi 4 marta ortganda, uning tebranishlar chastotasi qanday o'zgaradi?
A) 4 marta kamayadi B) 2 marta ortadi C) 4 marta ortadi D) 2 marta kamayadi
Berilgan:
π1 = 4π1 Yechilishi: Matematik mayatnikning tebranish chastotasini topish
π1
=? formulasidan foydalanam:
π2
π
Tenglamalar nisbatini olamiz:
π
Javob: D) 2 marta kamayadi.
28.6. Matematik mavatnikning yerdagi tebranish davri T ga teng bo' lsa, erkin tushish tezlanishg yerdagidan n marta katta boβlgan planetadagi tebranish davri qanday bo' ladi?
π
A) B)βππ C) π D)ππ
Berilgan:
ππ¦ππ = π Yechilishi: Matematik mayatnikning tebranish davrini topish ππ = πππ¦ππ formulasidan foydalanamiz
ππ =?
ππ¦ππ = 2π
Tenglama nisbatini olamiz.
= π = πππ¦ππ , π π π
ππ 2πβ π{π¦ππ
ππ
π
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