# Two different values for two different methods of calculating spring constant [closed]

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, $$k = 78.96$$

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

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• Where does $-kx=mg$ come from? – Acccumulation Jan 30 at 22:25
• @Acccumulation $F = ma = mg$ when the system is hanging. Since $F = -kx$, I set the two equal to each other. – Jay Jan 30 at 22:28
• Who said anything about hanging? – Pieter Jan 30 at 22:29
• 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". – Acccumulation Jan 30 at 22:31

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 $$\Sigma F = F_s + F_g = ma \neq 0.$$