Question

A 150 g air-track glider is attached to a spring. The glider is pushed in 8.80 cm and released. A student with a stopwatch finds that 10.0 oscillations take 17.0 s .What is the spring constant?

Answer 1

2.06 N/m

Step-by-step explanation:

The system makes 10.0 complete oscillations in 17.0 s. So, the frequency of the system is

`f=(10.0)/(17.0 s)=0.59 Hz`

The angular frequency of the system is given by

`\omega = 2\pi f=2\pi (0.59 Hz)=3.71 rad/s`

In a simple harmonic motion, the angular frequency is related to the mass and the spring constant by

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

where

k is the spring constant

m is the mass

Here we know

`\omega=3.71 rad/s\\m = 150 g = 0.150 kg`

So we can solve the formula to find k:

`k=\omega^2 m = (3.71 rad/s)^2 (0.150 kg)=2.06 N/m`