Question

A 190 g air-track glider is attached to a spring. The glider is pushed in 9.20 cm and released. A student with a stopwatch finds that 14.0 oscillations take 11.0 s. What is the spring constant?

Answer 1

Answer:

$k = 12.136\,\frac{N}{m}$

Explanation:

The angular frequency of the system is:

$\omega = \sqrt{\frac{k}{m} }$

The frequency is:

$f = \frac{14\,osc}{11\,s}$

$f = 1.272\,hz$

The angular frequency is:

$\omega = 2\pi\cdot (1.272\,hz)$

$\omega = 7.992\,\frac{rad}{s}$

The spring constant is:

$k = \left(7.992\,\frac{rad}{s} \right)^{2}\cdot (0.190\,kg)$

$k = 12.136\,\frac{N}{m}$