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Если смещение X задается уравнением X=Asin(ωt+α), (1) то скорость Vx=∨ = Aωcos(ωt+α) (2) Разделим уравнение (1) на A, а уравнение (2) на Aω, возведем в квадрат и получим, используя основное тригонометрическое тождество: (X/A)² + (v/Aω)²=1. Поэтому амплитуда
A=√X²+(v/ω)²=√25²+(100/4)²=25√2 см
Так как в условии задачи рассматривается момент времени, начиная от положения равновесия, то можно принять α=0 Поэтому, используя уравнение (1), получим X/A =1/√2=√2/2, а ωt=π/4, следовательно, t=π/(4ω)=π/(4×4)=π/16≈0,2
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X=Asin(ωt+α), (1)
то скорость
Vx=∨ = Aωcos(ωt+α) (2)
Разделим уравнение (1) на A,
а уравнение (2) на Aω, возведем в квадрат и получим, используя основное тригонометрическое тождество:
(X/A)² + (v/Aω)²=1.
Поэтому амплитуда
A=√X²+(v/ω)²=√25²+(100/4)²=25√2 см
Так как в условии задачи рассматривается момент времени, начиная
от положения равновесия, то можно принять
α=0
Поэтому, используя уравнение (1), получим
X/A =1/√2=√2/2, а
ωt=π/4,
следовательно,
t=π/(4ω)=π/(4×4)=π/16≈0,2
Объяснение:
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