Question

Материальная точка совершает гармоничекие колебания с амплитудой 5 см. Максимальная ускорение точки составляет 20 см/с^2. Определить циклическую частоту колебания точки.

Answer 1

a=0.2     A=0.05    w=?

a=A*w²      w=√(a/A)=√(0.2/0.05)=2 рад/с   Здесь а и А - максимальные значения ускорения и амплитуды.

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Answer 2

Calculation of the cyclic frequency of oscillation

To determine the cyclic frequency of the oscillation of the material point, we can use the formula:

Cyclic frequency (ω) = Maximum acceleration (a) / Amplitude (A)

Given that the maximum acceleration of the point is 20 cm/s^2 and the amplitude is 5 cm, we can substitute these values into the formula to calculate the cyclic frequency.

Cyclic frequency (ω) = 20 cm/s^2 / 5 cm

Converting the units to be consistent, we have:

Cyclic frequency (ω) = 20 cm/s^2 / 0.05 m

Simplifying the expression, we get:

Cyclic frequency (ω) = 400 rad/s

Therefore, the cyclic frequency of the oscillation of the material point is 400 rad/s.