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A 130g air-track glider is attached to a spring. The glider is pushed in 10.4cm and released. A student with a stopwatch finds that 14.0 oscillations take 19.0s

I would like to know why the answer I get is wrong. I set up an equation to represent position as a function of time.

$$x(t) = \frac{52}{5}*cos(\frac{19\pi*t}{7})$$

$$a(t) = \frac{d^2x(t)}{dt^2} = -\frac{18772\pi^2}{245}*cos(\frac{19\pi *t}{7})$$

Also the force of the spring $$F_s = -kx$$

$$\Sigma F_x = ma_x$$

So plug in $t = 0$ $s$ and $x = 10.4$ $cm$ for simplicity

Then the acceleration is the constant $-\frac{18772\pi^2}{245}$ and multiply by the mass to get the force. We also know the position is $10.4$ $cm$ when the force is such.

$$k = -\frac{m*a(t)}{x(t)} = - \frac{0.13 * (-\frac{18772\pi^2}{245})}{10.4} = 9.45_\frac{N}{cm}$$

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Remember that an oscillation can be written as $\cos(\omega t) = \cos(2\pi f t) = \cos(\frac{2 \pi}{T} t)$. You have the period on the top instead of the bottom.

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  • $\begingroup$ I calculated the period is $\frac{oscillations}{time} = \frac{14}{19}$. This tells me $\frac{2*\pi}{T} = \frac{2*\pi*19}{14} = \frac{19*\pi}{7}$ $\endgroup$
    – Leonardo
    Oct 29, 2013 at 3:51
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    $\begingroup$ @Leonardo: The period is the time required per oscillation, i.e. $\frac{time}{oscillations} = \frac{19}{14}$ $\endgroup$
    – pho
    Oct 29, 2013 at 3:58

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