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An oscillating block-spring system has a mechanical energy of 1.00 J, amplitude of 9.3 cm, and a...

Question:

An oscillating block-spring system has a mechanical energy of {eq}1.00\ \rm{J} {/eq}, amplitude of {eq}9.3\ \rm{cm} {/eq}, and a maximum speed of {eq}1.34\ \rm{m/s} {/eq}.

(a) Find the spring constant.

(b) Find the mass of the block.

(c) Find the frequency of oscillation.

Show work.

Spring:

Spring is a device employed to absorb vibrations produced by an unbalanced mechanical structure. The properties like the stiffness of the spring, strain energy stored in the spring, amplitude, and frequency are used to define the application area of the spring.

Answer and Explanation: 1


We are given the following data:

  • Energy stored in the block-sping system, {eq}E=1\ \text{J} {/eq}
  • Amplitude of spring, {eq}x=9.3\ \text{cm} {/eq}
  • Speed of the block, {eq}V=1.34\ \text{m/s} {/eq}


Question (a)


We are asked to calculate the spring contact, we can do so by using the following equation:

{eq}E=\dfrac{1}{2}kx^{2} {/eq}

  • k is the spring constant


Substituting values in the above equation, we have:

{eq}\begin{align} E&=\dfrac{1}{2}kx^{2}\\[0.3 cm] 1\ \text{J}&=\dfrac{1}{2}\times k\times(9.3\times10^{-2}\ \text{m})^{2}&\left [ \text{1 cm=10}^{-2}\ \text{m} \right ]\\[0.3 cm] 1\ \text{N.m}&=\dfrac{1}{2}\times k\times (9.3\times10^{-2}\ \text{m})^{2}\\[0.3 cm] k&\approx\boxed{231.2\ \text{N/m}} \end{align} {/eq}


Question (b)


We are asked to calculate the mass of the block, we can do so by using the following equation:

For an equilibrium system, the energy stored in the spring is equal to the Kinetic energy of the block, we have:

{eq}\begin{align} E&=\dfrac{1}{2}mV^{2}\\[0.3 cm] 1\ \text{J}&=\dfrac{1}{2}\times m\times(1.34\ \text{m/s})^{2}\\[0.3 cm] 1\ \text{N.m}&=\dfrac{1}{2}\times m\times(1.34\ \text{m/s})^{2}\\[0.3 cm] m&\approx\boxed{1.113\ \text{kg}} \end{align} {/eq}


Question (c)


The frequency of the oscillation:

We can calculate the frequency of the oscillation by using the following equation:

{eq}\begin{align} f&=\dfrac{1}{2\pi}\sqrt{\frac{k}{m}}\\[0.3 cm] &=\dfrac{1}{2}\sqrt{\frac{(231.2\ \text{N/m})}{(1.113\ \text{kg})}}\\[0.3 cm] &\approx\boxed{2.29\ \text{Hz}} \end{align} {/eq}


Learn more about this topic:

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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.


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