Question

An object–spring system moving with simple harmonic motion has an amplitude A.

(a) What is the total energy of the system in terms of k and A only?

E :

(b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this situation, using only the variables for the mass m, velocity v, spring constant k, and position x.

(c) Using the results of parts (a) and (b) and the conservation of energy equation, find the positions x of the object when its kinetic energy equals twice the potential energy stored in the spring. (The answer should be in terms of A only.)

x = ±

Answer 1

a) This is just the elastic energy

E = 1/2 * k * A²

E = kA² / 2

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b) here, just equate kinetic energy to elastic potential energy

1/2mv² = 1/2kx²

it is given that kinetic energy is twice the elastic potential energy.

so,

1/2mv² = 2 * 1/2kx²

1/2mv² = kx²

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c) using conservation of energy,

E = 2 * P.E + P.E

E = 3 * P.E

so,

kA² / 2  = 3/2kx²

kA²   = 3kx²

x² = A² / 3

x = +/- A / sqrt (3)