Question

A 170g air-track glider is attached to a spring. The glider is pushed in 9.40 cm

 and released. A student with a stopwatch finds that 8.00 oscillations take 19.0 s.

What is the spring constant?

Express your answer with the appropriate units.

Answer 1

Explanations & Calculations

• The periodic time is $\small T = \large\frac{19.0s}{8}\small= 2.38s$
• The angular frequency of this type of simple harmonic motion is $\small \omega=\large \sqrt{\frac{k}{m}}$
• By the relationship $\small \omega=\large\frac{2\pi}{T}$ you can find the spring constant.

$\qquad\qquad \begin{aligned} \small \omega^2&=\small \frac{k}{m}=\frac{4\pi^2}{T^2}\\ \small k&=\small \frac{4\pi^2m}{T^2}\\ &=\small \frac{4\pi^2\times0.170kg}{(2.38s)^2}\\ &=\small 1.18 \,Nm^{-1} \end{aligned}$

• The frequency or the period does not change how deep the glider is pressed as they are characteristics of the spring & the attached mass. As long as they are constant, $\small f$ and $\small T$ remain constant.