Question

Амплитуда колебаний пружинной породы составляет 0,1 м. 0,5 кг прикрепленной к нему массы Максимальное значение скорости тела 2 м/с. Определить жесткость пружины.

Answer 1

Ответ:

200 Н/м

Объяснение:

A=0.1 м    m=0.5 кг     vm=2 м/с   k=?

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vm=A*ω=A*√(k/m)

k=m*(vm/A)²=0.5*(2/0.1)²=200 Н/м

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Answer 2

Calculation of the Spring Constant

To determine the spring constant (also known as the stiffness or force constant) of the given spring, we can use the equation that relates the amplitude of oscillation, the mass attached to the spring, and the maximum velocity of the body:

k = (m * v^2) / (2 * A^2)

Where: - k is the spring constant - m is the mass attached to the spring - v is the maximum velocity of the body - A is the amplitude of oscillation

Given information: - Amplitude of oscillation (A) = 0.1 m - Mass attached to the spring (m) = 0.5 kg - Maximum velocity of the body (v) = 2 m/s

Substituting the given values into the equation, we can calculate the spring constant:

k = (0.5 kg * (2 m/s)^2) / (2 * (0.1 m)^2)

Simplifying the equation:

k = (0.5 kg * 4 m^2/s^2) / (2 * 0.01 m^2)

k = (2 kg * m^2/s^2) / (0.02 m^2)

k = 100 N/m

Therefore, the spring constant of the given spring is 100 N/m.