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

26.13. Bikirligi 125 N/m bo’lgan prujinaga osilgan yuk 8 s da 10 marta tebranmoqda .Yukning qandey?(𝜋²=10)
A)6 B)4 C)2 D)8
Berilgan: K=125 N/m
t=8s
N=5
_____
m=?
Yechilishi:Prujinali mayatnik uchun tebranish davri formulasidan foydalanamiz:
𝑘

T=2 , T=
𝑁
Ikkala tenglamani tenglashtiramiz:
𝑚 𝑡
2𝜋√ = 𝑘 𝑁
Ildizlarni chiqarish uchun tenglamaning ikki tomonini kvadratga oshiramiz:
𝑚 𝑡2

4𝜋2 𝑘 = 𝑁2
Bu ifodadan quydagini olamiz:
𝑘𝑡² 125 × 𝑔² 8000
𝑚 = 4 𝜋2𝑁2 = 4 × 10 × 102 = 4000 = 2𝑘𝑔 
Javob:C)2.
26.14. Bikirligi 100N/m bo’lgan prujinaga qandey (g) massali yuk osilganda, uning tebranish chastotasi 5Hz bo’ladi?
A)10 B)25 C)50 D)100
Berilgan:
k=100N/m
𝜈 = 5𝐻𝑧 𝜋² ≈ 10 m=?
Yechilishi:Prujinali mayatnik uchun tebranish davri formulasidan foydalanamiz:
𝜈
𝑚
2 𝜋√
𝑘
Ildizlarni yo’qotish uchun tenglamaning ikki tomonini kvadratga oshiramiz:

𝜈
2 = 4𝜋𝑘2𝑚
Bundan massani topamiz.
𝑘 100 100
𝑚 = 4 𝜋₂𝜈2 = 4 × 10 × 52 = 1000 = 0.1𝑘𝑔 = 100𝑔
Javob.D)100.
26.15. Prujinaga osilgan jisimning tebranishlar davri T.Agar yukning massasi 60 g ortsa , tebranishlar davri 2 marta ortdi.Jismning dastlabki massasi qandey (g)?

A)10 Berilgan:	B)20 C)40 D)60
Τ1 = Τ	Yechilishi:Prujinali mayatnik uchun tebranish davdri ∆𝑚 = 60𝑔 = 0.06𝑘𝑔 formulasidan foydalanamiz. 𝑚2 = 𝑚1 + ∆𝑚
Τ2 = 2Τ	𝑚1 =?

𝑚¹
𝑇₁ = 2𝜋√ _𝑘 , 𝑇₂
Nisbat olamiz:
2𝜋√ 1 1 1
𝑇1
; .
𝑇₂ 2𝑇 2 √𝑚₂ 2 𝑚₂ 4
Bunada 4𝑚₁ = 𝑚₂ = 𝑚₁ = ∆𝑚
4𝑚₁ − 𝑚₁ = ∆𝑚;
3𝑚₁ = ∆𝑚;
∆𝑚 0.06
𝑚₁ = = = 0.02𝑘𝑔 = 20𝑔
3 3