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A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a...

Question:

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall while the other end is connected to a movable object. The spring and object are compressed by 0.074 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 11.6 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.041 m relative to its unstrained length?

Spring:

The spring can be defined as a mechanical device that can store potential energy and release kinetic energy. Some types of springs are as follows: Compression Spring, Extension Spring, Torsion Spring, etc.

Answer and Explanation: 1


We are given the following data:

  • The angular speed is {eq}\omega = 11.6\;{\rm{rad/s}} {/eq}.
  • The spring and objects are compressed by {eq}A = 0.074\;{\rm{m}} {/eq}.
  • The stretch length of the spring is {eq}x = 0.042\;{\rm{m}} {/eq}.


The expression for the speed of the object at the instant is,

{eq}V = \omega \sqrt {\left( {{A^2} - {x^2}} \right)} {/eq}


Substitute the given values in the above expression.

{eq}\begin{align*} V &= 11.6\;{\rm{rad/s}}\sqrt {\left( {{{0.074}^2} - {{0.041}^2}} \right)} \\[0.3 cm] &= 11.6\sqrt {{\rm{0}}{\rm{.003795}}} \\[0.3 cm] &= {\rm{0}}{\rm{.7146}}\;{\rm{m/s}}\\[0.3 cm] &\approx {\rm{0}}{\rm{.715}}\;{\rm{m/s}} \end{align*} {/eq}


Thus, the speed of the object at the instant is {eq}\boxed{ {\rm{0}}{\rm{.715}}\;{\rm{m/s}}} {/eq}.


Learn more about this topic:

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Practice Applying Spring Constant Formulas

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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.


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