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## Homework Statement

A body of mass 4[kgr] is moving along the x axis while the following force is applied on it:

[tex] F= -3(x-6) [/tex]

We know that at time t=0 the kinetic energy is [tex] K=2.16[J] [/tex] and that its decreasing, that is, [tex] \frac{dK}{dt}<0 [/tex].

The potential energy (with respect to the equilibrium point) at that same instant is: [tex] V=0.96[J] [/tex]

When will this body return to its initial point (i.e. the position at t=0)?

**2. The attempt at a solution**

The initial position of the body with respect to the equilibrium point (x=6) is found from the P.E. expression, as the "spring" constant is k=3. So:

[tex]V=0.5kx^{2} \to x= \pm 0.8 [/tex]

Now, we find the position of the body with respect to time. The amplitude is found from energy conservation between the initial state and when the body is at a maximum distance from equilibrium (K=0), so that:

[tex] K(t=0)+V(t=0) = 0.5kA^{2} \to A=1.442 [/tex]

So,

[tex] x(t) = 1.442cos(\omega t + \phi) [/tex] where [tex] \omega^{2}=k/m=0.75 [/tex] and [tex]\phi[/tex] is found from the initial position of the body: [tex] x= \pm 0.8 [/tex]

for the +0.8 we get after substitution: [tex] \phi = 56.3^{o} [/tex]

for the -0.8 we get after substitution: [tex] \phi = 33.7^{o} [/tex]

The data of [tex] \frac{dK}{dt} < 0 [/tex] means that the body is drawing away from the initial position, so had it been in x=+0.8 then it moves to the right and had it been in in x=-0.8 then it moves to the left.

So I have everthing I need to know about this oscillator. My question: What is the condition I need in order to answer what is required, i.e., the condition of the return of the body to its initial place?

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