# A 170 g air-track glider is attached to a spring. The glider is pushed in 11.6 cm against the...

## Question:

A 170 kg air-track glider is attached to a spring. The glider is pushed in 11.6 cm against the spring, then released. A student with a stopwatch finds that 12 oscillations take 16.0 s. What is the spring constant?

## Time Period in SHM:

Time period of a particle undergoing simple harmonic motion is defined as the time taken by the particle to complete a full oscillation about its mean position.

Time period of oscillation of a mass m attached to a spring having force constant k is given by

\begin{align*} T&=\frac{1}{2\pi}\sqrt {\frac{m}{k}} \\ \end{align*}

## Answer and Explanation: 1

Given:

• Mass of the glider is m = 170 kg .
• The spring makes 12 oscillations take 16.0 s.

Time taken to make one oscillation is equal to the time period of oscillation of the glider is

{eq}T =\displaystyle \frac{16}{12} \ s {/eq}.

Let the spring constant be k.

Thus from the expression of time period we have

{eq}\begin{align*} T&=\frac{1}{2\pi}\sqrt {\frac{m}{k}} \\ \frac{16}{12} &=\frac{1}{2\pi}\sqrt {\frac{170}{k}} \\ k&=2.42 \ N/m\Rightarrow(Answer) \end{align*} {/eq}

#### Learn more about this topic: Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
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In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.