Question

Колебания точки заданы уравнением X=10cos1/3Пt. Найдите амплитуду, частоту, и период колебаний считая что все величины выражены в единицах СИ

Answer 1

из условия имеем: амплитуда-10.  Следовательно из формул:частота=2, период 0,5

Answer 2

Calculating Amplitude, Frequency, and Period of Oscillations

The equation given for the oscillations of a point is X = 10cos(1/3πt). Let's calculate the amplitude, frequency, and period of the oscillations.

Amplitude: The amplitude of the oscillation is the maximum displacement from the equilibrium position. In the given equation, the amplitude is 10.

Frequency: The frequency of the oscillation is the number of complete cycles per unit time. In the given equation, the frequency can be determined from the coefficient of t in the cosine function. The coefficient of t is 1/3π. The frequency can be calculated using the formula: f = ω/2π, where ω is the angular frequency. In this case, ω = 1/3π. Therefore, the frequency is f = (1/3π)/(2π) = 1/6.

Period: The period of the oscillation is the time taken to complete one full cycle. It is inversely proportional to the frequency. The period can be calculated using the formula: T = 1/f. Substituting the frequency calculated above, the period is T = 1/(1/6) = 6.

Therefore, the amplitude is 10, the frequency is 1/6, and the period is 6 when all quantities are expressed in SI units.