# A 100 coil spring has a spring constant of 438 N/m. It is cut into two shorter springs, each of...

## Question:

A 100 coil spring has a spring constant of 438 N/m. It is cut into two shorter springs, each of which has 50 coils. One end of a 50-coil spring is attached to a wall. An object of mass 37 kg is attached to the other end of the spring, and the system is set into horizontal oscillation. What is the angular frequency of the motion?

## Spring Constant:

When spring is cut into a certain specified number of parts, then the spring constant of a smaller spring is greater than the original spring constant. As the number of cuts of a spring increases, then the new spring constant of shorter spring also increases. There is an inverse relationship between the spring constant and length.

## Answer and Explanation: 1

**Given Data**

- The spring constant is {eq}k = 438\;{\rm{N/m}} {/eq}.

- The mass of an object is {eq}m = 37\;{\rm{kg}} {/eq}.

As we know one thing, when a spring with a specified number of a coil is cut into two equal parts then the spring constant of any smaller parts is equal to double of the spring constant of the original long spring. So the formula to calculate the spring constant of smaller spring is given by,

{eq}{k_{{\rm{small}}}} = 2{k_{{\rm{original}}}} {/eq}.

Substitute the known value in the above formula.

{eq}\begin{align*} {k_{{\rm{small}}}} &= 2{k_{{\rm{original}}}}\\ &= 2 \times 438\\ &= 876\;{\rm{N/m}} \end{align*} {/eq}

The formula to calculate the angular frequency of the motion is given by,

{eq}\omega = \sqrt {\dfrac{{{k_{{\rm{small}}}}}}{m}} {/eq}.

Here, {eq}\omega {/eq} is the angular frequency of motion.

Substitute all the known values in the above formula.

{eq}\begin{align*} \omega &= \sqrt {\dfrac{{{k_{{\rm{small}}}}}}{m}} \\ &= \sqrt {\dfrac{{876}}{{37}}} \\ &\approx 4.87\;{\rm{rad/s}} \end{align*} {/eq}

Thus, the angular frequency of motion is {eq}4.87\;{\rm{rad/s}} {/eq}.

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Chapter 17 / Lesson 11In 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.