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