The amplitude of oscillations of a spring pendulum with a mass of 100 g and a stiffness of 40 N \ m is 5 cm. Find the speed of a body when displaced by 3 cm.
m = 100 g = 0.1 kg.
k = 40 N / m.
A = 5cm = 0.05m.
x = 3 cm = 0.03 m.
To solve the problem, we will use the law of conservation of total mechanical energy: En = En1 + Ek1.
En = k * A ^ 2/2.
En1 = k * x ^ 2/2.
Ek1 = m * V ^ 2/2.
k * A ^ 2/2 = k * x ^ 2/2 + m * V ^ 2/2.
V2 = k * A ^ 2 / m – k * x2 / m.
The speed of the load will be determined by the formula: V = √ (k * A ^ 2 / m – k * x ^ 2 / m).
V = √ (40 N / m * (0.05 m) ^ 2 / 0.1 kg – 40 N / m * (0.03 m) ^ 2 / 0.1 kg) = 0.8 m / s.
Answer: the weight of the spring pendulum will have a speed of V = 0.8 m / s.
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