Question

Груз массой 200 г совершает колебания на пружине с жёсткостью 500 Н/м. Найдите частоту колебаний и наибольшую скорость движения груза, если амплитуда колебаний 8 см.

Answer 1

V=X*омегаомега = 2pi/T=2pi/2pi sqrt(m/k)= 1/0.02=50 рад/сВсе,теперь подставляемV = 8*10^-2*50=4 м/с

Answer 2

Calculation of Frequency and Maximum Velocity

To find the frequency of oscillation and the maximum velocity of the mass, we can use the following formulas:

Frequency (f) = 1 / (2π) * √(k / m)

Maximum Velocity (v_max) = A * 2π * f

where: - k is the spring constant (stiffness) in N/m, - m is the mass of the object in kg, - A is the amplitude of oscillation in meters, - π is a mathematical constant approximately equal to 3.14159.

Given: - Mass (m) = 200 g = 0.2 kg - Spring constant (k) = 500 N/m - Amplitude (A) = 8 cm = 0.08 m

Let's calculate the frequency and maximum velocity.

Step 1: Convert the mass from grams to kilograms: - Mass (m) = 200 g = 0.2 kg

Step 2: Calculate the frequency (f): - Frequency (f) = 1 / (2π) * √(k / m) - Frequency (f) = 1 / (2π) * √(500 / 0.2) - Frequency (f) ≈ 1 / (2π) * √(2500) - Frequency (f) ≈ 1 / (2π) * 50 - Frequency (f) ≈ 0.0796 Hz

Step 3: Calculate the maximum velocity (v_max): - Maximum Velocity (v_max) = A * 2π * f - Maximum Velocity (v_max) = 0.08 * 2π * 0.0796 - Maximum Velocity (v_max) ≈ 0.08 * 2 * 3.14159 * 0.0796 - Maximum Velocity (v_max) ≈ 0.0503 m/s

Therefore, the frequency of oscillation is approximately 0.0796 Hz, and the maximum velocity of the mass is approximately 0.0503 m/s.

Please note that the calculations were performed based on the given information and formulas.