Question

Решить задачу: тело массой 500г. совершает колебания с на пружине жёсткостью 32Н/м, амплитуда колебаний равна 5см. Определите циклическую частоту данных колебаний.

Answer 1

Ответ:

8 рад/с

Объяснение:

m = 500 г = 0,5 кг

k = 32 Н/м

x(m) = 5 cм = 0,05 м

ω - ?

=========

ω = √(k/m) = √(32/0,5) = √64 = 8 рад/с

Answer 2

Calculation of the Circular Frequency of Oscillations

To determine the circular frequency of the given oscillations, we can use the formula:

ω = √(k/m)

where: - ω is the circular frequency (in radians per second) - k is the spring constant (in Newtons per meter) - m is the mass of the object (in kilograms)

In this case, the mass of the object is given as 500 grams, which is equal to 0.5 kilograms. The spring constant is given as 32 N/m.

Substituting these values into the formula, we have:

ω = √(32/0.5)

Calculating this expression, we find:

ω ≈ 10.67 rad/s

Therefore, the circular frequency of the given oscillations is approximately 10.67 radians per second.