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

`k = 12.136\,(N)/(m)`

Step-by-step explanation:

The angular frequency of the system is:

`\omega = \sqrt{(k)/(m) }`

The frequency is:

`f = (14\,osc)/(11\,s)`

`f = 1.272\,hz`

The angular frequency is:

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

`\omega = 7.992\,(rad)/(s)`

The spring constant is:

`k = \left(7.992\,(rad)/(s) \right)^(2)\cdot (0.190\,kg)`

`k = 12.136\,(N)/(m)`