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I'm given the question: "An oscillator consists of a block of mass .5 kg connected to a spring. When set into oscillation with amplitude .35 m, the oscillator repeats its motion every .5 seconds. Find its spring constant."

Approach 1: Force = $-kx = mg$. It follows then that $k = \frac{-mg}{x}$.

$k = \frac{-(.5)(9.81)}{.35} = 14.0$

Approach 2: I know that $T=2\pi\sqrt{\frac{m}{k}}$. It follows then that $.5 = 2\pi\sqrt{\frac{.5}{k}}$. Solving for k, $k = 78.96$

Why are these answers different?

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closed as off-topic by G. Smith, Gert, John Rennie, Jon Custer, Buzz Jan 31 at 20:33

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    $\begingroup$ Where does $-kx=mg$ come from? $\endgroup$ – Acccumulation Jan 30 at 22:25
  • $\begingroup$ @Acccumulation $F = ma = mg$ when the system is hanging. Since $F = -kx$, I set the two equal to each other. $\endgroup$ – Jay Jan 30 at 22:28
  • $\begingroup$ Who said anything about hanging? $\endgroup$ – Pieter Jan 30 at 22:29
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    $\begingroup$ You didn't include any mention in the problem of it hanging. And it's a fallacy to say "X is a force, Y is a force, therefore X is equal to Y". $\endgroup$ – Acccumulation Jan 30 at 22:31
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In your question, Approach #1 is invalid. You implicitly assumed that $$ \Sigma F = F_s + F_g = 0. $$

This, however, is not the case. When the mass is oscillating vertically (as I assume you mean for it to be), the the acceleration is nonzero, and as such $$ \Sigma F = F_s + F_g = ma \neq 0. $$

Thus, you should approach this problem either from the standpoint of energy, or from the standpoint of a simple harmonic oscillator, as you have done in approach 2.

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