Question

3. Consider a mass m attached to a horizontal spring with spring constant k. Suppose the mass is pulled a distance A from the equilibrium position.

a. Find the total energy at the amplitude in terms of k and A

b. Using conservation of energy, find an expression for the maximum speed at the equilibrium position in terms of k, A and m

Answer #1

given m = mass , k = spring constant , X = A = distance from equilibrium position

(a) total energy at the amplitude = kinetic energy + spring
potential energy = 0 + (1/2) k A^{2} =
0.5kA^{2}

(b) applying energy conservation principle

total energy at equilibrium position = total energy at stretched position

=> Kinetic energy at equilibrium + spring potential energy = kinetic energy at stretched position + spring potential energy at stretched position

=> 0.5mV^{2} + 0 = 0 + 0.5kA^{2}

=> V^{2} = (kA^{2}) / m

=>V = m/s

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