Question

Тіло здійснює коливання за законом х = 0,2 cos4t. Знайдіть 1) амплітуду 2) період 3) частоту коливань. Тело совершает колебания по закону х = 0,2 cos4t. Найдите 1) амплитуду 2) период 3) частоту колебаний.

Answer 1

Ответ:

Объяснение:

х = 0,2 cos4t

или

x=Acoswt

А - амплитуда ( по системе СИ в метрах )

w - циклическая частота

$w=2\pi v$  где $v$ - частота колебаний

А = 0,2 метра

$v=\frac{w}{2\pi } =\frac{4}{2\pi }=\frac{2}{\pi }=0,64$ Гц (считал $\pi$ как 3.14)

$T=\frac{1}{v}=\frac{100}{64} = 1,5625 c$

Answer 2

Body oscillates according to the law x = 0.2 cos(4t). Find: 1) amplitude 2) period 3) frequency of oscillations.

To find the amplitude, period, and frequency of the oscillations described by the equation x = 0.2 cos(4t), we can use the following formulas:

1) Amplitude (A): The amplitude of an oscillation is the maximum displacement from the equilibrium position. In this case, the amplitude is 0.2.

2) Period (T): The period of an oscillation is the time it takes for one complete cycle. It can be calculated using the formula T = 2π/ω, where ω is the angular frequency. In this case, the angular frequency is 4, so the period can be calculated as T = 2π/4 = π/2.

3) Frequency (f): The frequency of an oscillation is the number of complete cycles per unit of time. It can be calculated using the formula f = 1/T. In this case, the period is π/2, so the frequency can be calculated as f = 1/(π/2) = 2/π.

Therefore, the answers to the given questions are: 1) The amplitude is 0.2. 2) The period is π/2. 3) The frequency is 2/π.

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