ﻣﺴﺎﻟﺔ ﺷﻬﺎﺩﺓ 2009

ﻣﺴﺎﻟﺔ ﺷﻬﺎﺩﺓ 2009 ﻭ ﺣﻠﻬﺎ
ﺟﺴﻢ ﻳﺘﺤﺮﻙ ﺣﺮﻛﺔ ﺗﻮﺍﻓﻘﻴﺔ ﺑﺴﻴﻂ . ﺍﺫﺍ
ﻛﺎﻥ ﺍﻻﺗﺴﺎﻉ 0.01 ﻣﺘﺮ ﻭ ﺗﺮﺩﺩ 10 ﻫﻴﺮﺗﺰ
ﺃﻭﺟﺪ :
– 1 ﻣﻌﺎﺩﻟﺔ ﺣﺮﻛﺔ ﺍﻟﺠﺴﻢ
– 2 ﺍﻟﺴﺮﻋﺔ
– 3 ﺍﻟﻌﺠﻠﺔ ﺑﻌﺪ ﺛﺎﻧﻴﺘﻴﻦ
ﺍﻟﺤﻞ :
- 1 ﻣﻌﺎﺩﻟﺔ ﺣﺮﻛﺔ ﺍﻟﺠﺴﻢ
ﺍﻟﻤﻌﻄﻴﺎﺕ : ﺃ = 0.01 ﺫ 10 = ﻫﻴﺮﺗﺰ
ﺹ = ﺃ ﺟﺎ ω ﻥ
= ω ﺫ π 2 ×
π2 × 10 = 20 π
ﻣﻌﺎﺩﻟﺔ ﺣﺮﻛﺔ ﺍﻟﺠﺴﻢ ﺹ = 0.01 ﺟﺎ 20
π ﻥ
– 2 ﺍﻟﺴﺮﻋﺔ
ﻉ = ﺃ ω ﺟﺘﺎ ω ﻥ
ﻉ = 0.01 π 20 × ﺟﺘﺎ π 20 ﻥ
ﻉ = 0.2 π ﺟﺘﺎ π 20 ﻥ
ﻉ = 0.2 π ﻣﺘﺮ / ﺙ
– 3 ﺍﻟﻌﺠﻠﺔ ﺑﻌﺪ ﺛﺎﻧﻴﺘﻴﻦ
ﺟـ - = ﺃ ω2 ﺟﺎ ω ﻥ
ﺟـ - = 0.01 400 × π2 ﺟﺎ π 20 ﻥ
ﺟـ 4 - = π2 ﺟﺎ π 20 ﻥ
ﻋﻨﺪ ﻥ 2 =
ﺟـ 4 - = π2 ﺟﺎ 20 π × 2
ﺟـ 4 - = π2 ﺟﺎ 40π = ﺻﻔﺮ ﻡ / ﺙ2
ﻣﻠﺤﻮﻇﺔ ) π2 ﺗﻌﻨﻰ ﺑﺎﻯ ﺗﺮﺑﻴﻊ ﻭ ω2
ﺗﻌﻨﻰ ﺍﻭﻣﻴﻘﺎ ﺗﺮﺑﻴﻊ (