Question

A $200g$air-track glider is attached to a spring. The glider is pushed in 10 cm .and released. A student with a stopwatch finds that $10$oscillations take 12.0 s . What is the spring constant?

Short answer

The spring constant of air track glider is$k=5.48\mathrm{N}/\mathrm{m}$

Step-by-step solution

Step :1 Given data 

Mass of the glider is 0.200 g .

No of oscillations is 10 .

Time taken is 12.0 s

$m=0.2\mathrm{kg}$

$N=10cycles$

$t_N=12\mathrm{s}$

Required :$K$

$A=10.0cm$

Step :2  Magnitude  of time period

The amount of time it takes to complete one cycle, also known as periodic time, is given by

$T=\frac{2\pi }{\omega }$=$2\pi \sqrt{\frac{m}{k}}$

localid="1650301186748" $$\begin{gathered} T=\frac{12}{10} \\ =1.2\mathrm{s} \end{gathered}$$

$\omega$is the angular frequency. k is force constant and m is mass attached with spring.

$$\begin{gathered} 1.2=2\pi \sqrt{\frac{.200}{k}} \\ 1.44=39.4\times \frac{.2}{k} \\ k=5.47N{m}^{-1} \end{gathered}$$