ﺘﻤﺎﺭﻴﻥ ﺤﻭل ﺍﻝﻤﻭﺠﺎﺕ ﺍﻝﻤﺘﻭﺍﻝﻴﺔ ﺍﻝﺠﻴﺒﻴﺔ (Iﻨﻌﺘﺒﺭ ﺍﻝﺘﺭﻜﻴﺏ ﺍﻝﺘﺎﻝﻲ: ﻗﻁﻥ ﻻﻤﺘﺼﺎﺹ ﺍﻝﻤﻭﺠﺔ ﺍﻝﻭﺍﺭﺩﺓ ≡ S ﻴﺤﺩﺙ ﺍﻝﻬﺯﺍﺯ ﻤﻭﺠﺔ ﻤﺘﻭﺍﻝﻴﺔ ﺩﻭﺭﻴﺔ ﺠﻴﺒﻴﺔ ﻁﻭل ﺍﻝﺤﺒل ، υ = 200 Hzﺴﺭﻋﺔ ﺍﻨﺘﺸﺎﺭﻫﺎ . v = 40m / s . ﻨﻀﻲﺀ ﺍﻝﺤﺒل ﺒﻭﺍﺴﻁﺔ ﻭﻤﺎﺽ ﺘﺭﺩﺩ ﻭﻤﻀﺎﺘﻪ υ eﺘﺭﺩﺩﻫﺎ (1ﺍﺤﺴﺏ ﻁﻭل ﺍﻝﻤﻭﺠﺔ . λ (2ﻤﺎ ﺃﻜﺒﺭ ﻗﻴﻤﺔ ﻝﺘﺭﺩﺩ ﺍﻝﻭﻤﺎﺽ ﺍﻝﺘﻲ ﺘﻤﻜﻥ ﻤﻥ ﻤﺸﺎﻫﺩﺓ ﺍﻝﺘﻭﻗﻑ ﺍﻝﻅﺎﻫﺭﻱ؟ (3ﻋﻠﻤﺎ ﺃﻥ ﺸﻜل ﺍﻝﻤﻭﺠﺔ ﺍﻝﻤﺘﻭﺍﻝﻴﺔ ﻫﻭ ﻜﻤﺎ ﻴﻠﻲ: ﺃ(ﻤﺎ ﺍﻝﺴﻠﻡ ﺍﻝﺫﻱ ﺘﻡ ﺍﺴﺘﻌﻤﺎﻝﻪ ﻓﻲ ﻫﺫﻩ ﺍﻝﻭﺜﻴﻘﺔ؟ ﺏ( ﻭﻀﺢ ﻜﻴﻔﻴﺔ ﺍﻫﺘﺯﺍﺯﺍﻝﻤﻨﺒﻊ Sﻋﻨﺩ ﺍﻝﻠﺤﻅﺔ . t = o ﺝ( ﻤﺜل ﻤﻅﻬﺭ ﺍﻝﺤﺒل ﺒﺎﻝﺴﻠﻡ ﺍﻝﺴﺎﺒﻕ ﻓﻲ ﺍﻝﻠﺤﻅﺔ . t = 0,015s ﺩ( ﺍﻋﻁ ﺘﻌﺒﻴﺭ ﺍﺴﺘﻁﺎﻝﺔ ﺍﻝﻤﻨﺒﻊ Sﺒﺩﻻﻝﺔ ﺍﻝﺯﻤﻥ ﻋﻠﻤﺎ ﺃﻨﻬﺎ ﻤﻨﻌﺩﻤﺔ ﻋﻨﺩ ﺍﻝﻠﺤﻅﺔ . t = o (4ﻨﻀﺒﻁ ﺍﻝﻭﻤﺎﺽ ﻋﻠﻰ ﺍﻝﺘﺭﺩﺩ .198 Hz ﺃ(ﺒﺎﺴﺘﻌﻤﺎل ﻫﺫﺍ ﺍﻝﺘﺭﺩﺩ ﻤﺎﺫﺍ ﻨﺸﺎﻫﺩ؟ﻋﻠل ﺠﻭﺍﺒﻙ ﺏ( ﺍﺤﺴﺏ ﺍﻝﻤﺴﺎﻓﺔ ﺍﻝﺘﻲ ﻗﻁﻌﺘﻬﺎ ﺍﻝﻤﻭﺠﺔ ﺨﻼل ﺍﻝﻤﺩﺓ ﺍﻝﺯﻤﻨﻴﺔ ﺍﻝﻔﺎﺼﻠﺔ ﺒﻴﻥ ﻭﻤﻀﺘﻴﻥ ﻤﺘﺘﺎﻝﻴﺘﻴﻥ . ﺝ( ﻋﻠﻤﺎ ﺃﻥ ﺘﺭﺩﺩ ﺍﻝﺤﺭﻜﺔ ﺍﻝﻅﺎﻫﺭﻴﺔ ﺍﻝﺒﻁﻴﺌﺔ ﻫﻭ Na = N − Ne ﺍﺴﺘﻨﺘﺞ ﺍﻝﺴﺭﻋﺔ ﺍﻝﻅﺎﻫﺭﻴﺔ ﻝﺤﺭﻜﺔ ﺍﻝﻤﻭﺠﺔ. (5ﺼﻑ ﺍﻝﻅﺎﻫﺭﺓ ﺍﻝﻤﺸﺎﻫﺩﺓ ﻋﻨﺩ ﻀﺒﻁ ﺍﻝﻭﻤﻀﺎﺕ ﻋﻠﻰ ﺍﻝﻘﻴﻤﺔ 202 Hz :ﻋﻠل ﺠﻭﺍﺒﻙ. ////////////////////////////////////////////////////////////////////////////////////////////////////// 40m / s (1ﻁﻭل ﺍﻝﻤﻭﺠﺔ = 0,2m = 20cm : 200 s −1 = v υ =λ --------------------------------------------------- (2ﺍﻝﻌﻼﻗﺔ ﺒﻴﻥ ﺘﺭﺩﺩ ﺍﻝﻭﻤﺎﺽ ﻭﺘﺭﺩﺩ ﺍﻫﺘﺯﺍﺯ ﺍﻝﺸﻔﺭﺓ ﻝﻤﺸﺎﻫﺩﺓ ﺍﻝﺘﻭﻗﻑ ﺍﻝﻅﺎﻫﺭﻱ ﻫﻲ: N = kNe ﺃﻱ: N k = Neﻤﻊ * k ∈ Ζ k =1 ﻭﺃﻜﺒﺭ ﺘﺭﺭﺩ ﻝﻠﻭﻤﺎﺽ ﺍﻝﺫﻱ ﻴﻤﻜﻥ ﻤﻥ ﻤﺸﺎﻫﺩﺓ ﺍﻝﺘﻭﻗﻑ ﺍﻝﻅﺎﻫﺭﻱ ﻴﻭﺍﻓﻕ ﺃﻱN = Ne = 200 Hz : ------------------------------------------------------- (3ﺃ( ﻤﻥ ﺨﻼل ﺸﻜل ﺍﻝﻤﻭﺠﺔ ،ﻨﻼﺤﻅ ﺃﻥ ﻁﻭﻝﻬﺎ ﻴﻭﺍﻓﻕ ﺃﺭﺒﻊ ﺴﻨﺘﻤﺘﺭﺍﺕ ﻋﻠﻰ ﺍﻝﻭﺜﻴﻘﺔ ﻭﻫﻭ ﻓﻲ ﺍﻝﻭﺍﻗﻊ ﻋﺸﺭﻭﻥ ﺴﻨﺘﻤﺘﺭ. 20cmﻤﻤﺜﻠﺔ ﻋﻠﻰ ﺍﻝﻭﺜﻴﻔﺔ ﺏ4cm : 1 ﺇﺫﻥ ﺍﻝﺴﻠﻡ ﻫﻭ :ﺃﻱ ﻜل cmﻋﻠﻰﺍﻝﺜﻴﻘﺔ ﻴﻤﺜل 5cm 5 --------------------------------------------------------- ﺏ(ﻤﻥ ﺨﻼل ﺸﻜل ﻤﻁﻠﻊ ﺍﻝﻤﻭﺠﺔ ﺍﻝﻤﺤﺫﺏ ﻴﺘﻀﺢ ﺃﻥ ﺍﻝﺸﻔﺭﺓ ﻋﻨﺩ ﺍﻝﻠﺤﻅﺔ t = 0ﺘﻬﺘﺯ ﻨﺤﻭ ﺍﻷﻋﻠﻰ. t T ﻝﺘﻤﺜﻴل ﻤﻅﻬﺭ ﺍﻝﺤﺒل ﻓﻲ ﻝﺤﻅﺔ tﻨﺤﺩﺩ ﻗﻴﻤﺔ ﺤﺎﺼل. : 1 ﻝﺩﻴﻨﺎ :ﺍﻝﺩﻭﺭ = 5 × 10 −3 s 200 = 1 υ =T t 0,015 = ﺇﺫﻥ= 3 : T 0,005 t = 0,015s ﻭﻤﻨﻪ ﻤﻅﻬﺭ ﺍﻝﺤﺒل ﻓﻲ ﺍﻝﻠﺤﻅﺔ ﺃﻱt = 3T : 3ﺃﺩﻭﺍﺭﻤﻊ ﺍﺤﺘﺭﺍﻡ ﺍﻝﻤﻁﻠﻊ ﺍﻝﻤﺤﺫﺏ ﻭﺍﻝﺫﻱ ﻴﺘﻌﻠﻕ ﺒﺎﻝﺤﺭﻜﺔ ﺍﻝﺒﺩﺌﻴﺔ ﻝﻠﺸﻔﺭﺓ ﺍﻝﻤﻬﺘﺯﺓ. ----------------------------------------------------------------------------------------------------------------ﺩ( ﺘﻌﺒﻴﺭ ﺇﺴﺘﻁﺎﻝﺔ S 2π 1 ) YS = A cos( t + ϕﻤﻥ ﺨﻼل ﺸﻜل ﺍﻝﻭﺠﺔ ﻝﺩﻴﻨﺎ ﻭﺒﺎﻋﺘﺒﺎﺭ ﺍﻝﺴﻠﻡ 5 T 1 1 = =T = 0,005s υ 200 إذن: A = 0,5 × 5 = 2,5cm 2π (YS = 2,5 × 10 − 2 cos )t 5 × 10 −3 ************************************************************************** &ن ه$ا ادد ا ! "#دد ا اا. ----------------------------------------------------------------------------------------------ب( ﺍﻝﻤﺴﺎﻓﺔ ﺍﻝﺘﻲ ﻗﻁﻌﺘﻬﺎ ﺍﻝﻤﻭﺠﺔ ﺨﻼل ﺍﻝﻤﺩﺓ ﺍﻝﺯﻤﻨﻴﺔ ﺍﻝﻔﺎﺼﻠﺔ ﺒﻴﻥ ﻭﻤﻀﺘﻴﻥ ﻤﺘﺘﺎﻝﻴﺘﻴﻥ: 40m / s = 0,202m = 20,2cm 198 Hz = v υe == d = v × Te -----------------------------------------------------------ج( دد ا0آ ا-.هـ:+ اإذن 23ا0آ ا-.ه +ه:1 Na = N − Ne = 200 − 198 = 2 Hz v a = d × υ a = 0,202 × 2 = 0,404m / s أي : v a = 40,4cm / s Vitesse du mouvement apparent : va -------------------------------------------------------------------------------------------------------------------------------------------------------------------- 89 :2(5ا-ض 62اددυ e = 202 Hz : -BCه; :آ -Aه; 60 <=2 1> ?@! +آ ا اا. إن ء ا Abdelkrim SBIRO )(Pour toutes observations contactez mon émail [email protected]