Question

Груз массой 100 грамм совершает колебания на пружине жесткостью 400 Н/м , имея амплитуду колебания 20 сантиметров . Найти максимальную скорось груза .

Answer 1

Сначала находим период
T=2pi√(m/k)

а теперь 
V=A*(2pi/T)=$A* \frac{2pi* \sqrt{ \frac{m}{k} } }{2pi} =A* \sqrt{ \frac{m}{k} }$=0,2*√(0,1/400)=.........

дальше уже калькулятор в помощь

Answer 2

Calculation of Maximum Velocity

To find the maximum velocity of a mass undergoing oscillations on a spring, we can use the formula:

v_max = A * ω

Where: - v_max is the maximum velocity - A is the amplitude of the oscillation - ω is the angular frequency of the oscillation

In this case, the mass has a mass of 100 grams (0.1 kg) and is undergoing oscillations on a spring with a stiffness of 400 N/m. The amplitude of the oscillation is 20 centimeters (0.2 meters).

To find the angular frequency, we can use the formula:

ω = √(k/m)

Where: - k is the stiffness of the spring - m is the mass of the object

Let's calculate the maximum velocity step by step.

1. Convert the mass to kilograms: - 100 grams = 0.1 kg

2. Calculate the angular frequency: - ω = √(k/m) - ω = √(400 N/m / 0.1 kg) - ω = √(4000 rad/s^2) - ω ≈ 63.25 rad/s

3. Calculate the maximum velocity: - v_max = A * ω - v_max = 0.2 m * 63.25 rad/s - v_max ≈ 12.65 m/s

Therefore, the maximum velocity of the mass undergoing oscillations on the spring is approximately 12.65 m/s.