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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1 Answer
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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 $x(t)=0.05\cos\left(\sqrt{\frac{k}{m}}*t\right) = \boxed{0.05\cos\left(10t\right)}$.