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A 2 kg block is attached to a spring for which $k=200N/m$ .

It is held at an extension of 5 cm and then released at t=0 ,

Find a, the displacement as a function of time and b, the velocity when $x ={\pm}a/2$

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The differential equation for an un-damped (amplitude remains constant and does not decrease over time) Simple Harmonic oscillator is

$\frac{\partial^2y}{\partial x^2} = -\left(\frac{k}{m}\right)x$

Meaning that the acceleration as a function of position is the negative of the spring constant over the mass.

Anyway, I am assuming the block is released from rest at 5 cm (0.05 m).

(A). The position formula in this case is going to be $x(t)=0.05\cos\left(\sqrt{\frac{k}{m}}*t\right) = \boxed{0.05\cos\left(10t\right)}$.

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