There are two springs. Their spring constants are related by k1 greater than k2. When identical...


There are two springs. Their spring constants are related by {eq}k_1 > k_2 {/eq}. When identical masses are oscillating on them, which one has a greater oscillation period?


A spring always tries to retain its equilibrium position. Therefore if you take a spring-mass system and disturb them from their equilibrium position then it will start to oscillate to retain its equilibrium position. This problem is based on the time period of that oscillation.

Answer and Explanation: 1

When a spring of spring constant {eq}k {/eq} is made to oscillate a mass {eq}m {/eq} then the period of the oscillation is given by $$\begin{align} T=2\pi\sqrt{\dfrac mk} \end{align} $$

When the mass is constant then this relation gives us:

$$\begin{align} T\propto \dfrac1{\sqrt k} \end{align} $$

This relation implies that the spring with a higher spring constant will have less time period.

From the question we have {eq}k_1\gt k_2 {/eq}, hence, we can conclude that the spring with {eq}k_2 {/eq} spring constant will have greater oscialltion period than the spring with {eq}k_1 {/eq} spring constant.

Learn more about this topic:

Practice Applying Spring Constant Formulas


Chapter 17 / Lesson 11

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.

Related to this Question

Explore our homework questions and answers library