At what speed does the weight of a spring pendulum, having a mass of 0.1 kg, pass through the equilibrium position if the stiffness of the spring is 40H \ m, and the amplitude of oscillation is 2 cm?
m = 0.1kg,
k = 40N / m,
X = 2cm = 0.02m;
Find: v -?
According to the law of conservation of energy, the potential energy of the spring at the maximum deviation of the load from equilibrium is equal to the kinetic energy of the load when passing through the equilibrium point:
Ep = Ek,
Ep = 1/2 * k * X ^ 2,
Ek = 1/2 * m * v ^ 2;
Substitute these expressions into the equation:
1/2 * m * v ^ 2 = 1/2 * k * X ^ 2;
Let’s find the required speed of the load:
v = X * sqrt (k / m) = 0.02 * sqrt (400) = 0.4m / s.
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