Ciało wykonuje drgania harmoniczne o amplitudzie A. Gdy jego wychylenie x=3/4A to Ek/Ep wynosi:
Witaj :)
dane: x=¾A
szukane: Ek/Ep
-------------------------------------------
Epmax = Ekmax = Ec = 0,5k*A².....energia całkowita Ec
Ep = 0,5k*x² = 0,5k*[¾A]² = [9/16]*0,5*k*A² = [9/16]*Ec
Ep = [9/16]*Ec
Ek = Ec - Ep = Ec - [9/16]*Ec = [7/16]*Ec
Ek = [7/16]*Ec
Ek/Ep = 7/9
Semper in altum..................................pozdrawiam :)
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Witaj :)
dane: x=¾A
szukane: Ek/Ep
-------------------------------------------
Epmax = Ekmax = Ec = 0,5k*A².....energia całkowita Ec
Ep = 0,5k*x² = 0,5k*[¾A]² = [9/16]*0,5*k*A² = [9/16]*Ec
Ep = [9/16]*Ec
Ek = Ec - Ep = Ec - [9/16]*Ec = [7/16]*Ec
Ek = [7/16]*Ec
Ek/Ep = 7/9
Semper in altum..................................pozdrawiam :)