A 1.0-kg block oscillates with a frequency of 10 Hz at the end of a certain spring. The spring is...

Question:

A 1.0-kg block oscillates with a frequency of 10 Hz at the end of a certain spring. The spring is then cut into two halves. The 1.0-kg block is then made to oscillate at the end of one of the halves. What is the frequency of oscillation of the block?

Spring-mass System:

It is very common to experience that a longer spring (more number of coils) can be easily stretched or compressed than the short spring. Even if the matter of the spring, the thickness of the wire remains constant, the spring constant increases with a decrease in the free length. Therefore, for the same hanging mass, the oscillation frequency of the block for a halved spring is higher than that of an original spring.

Answer and Explanation: 1


Given data:

  • {eq}m=\rm 1.0 \ kg {/eq} is the mass of the block
  • {eq}k_1 {/eq} is the spring constant of the original spring
  • {eq}k_2 {/eq} is the spring constant of one of the halves
  • {eq}f_1=\rm 10 \ Hz {/eq} is the initial frequency of oscillation
  • {eq}f_2 {/eq} is the frequency of oscillation of the block attached to one of the halves


When you split the spring in the middle, the stiffness of each piece becomes double. Therefore:

{eq}k_2=2k_1 {/eq}


The frequency of oscillation is given by:

{eq}f=\dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} {/eq}


The mass of the given block m remains constant. Therefore:

{eq}f \propto \sqrt {k} {/eq}


Using this mathematical relation, we write:

{eq}\begin{align} \dfrac{f_2}{f_1}&=\sqrt{\dfrac{k_2}{k_1}} \\[0.3cm] f_2&=f_1 \times \sqrt{\dfrac{2k_1}{k_1}} \\[0.3cm] f_2&=\text{ 10 Hz} \times \sqrt{\dfrac{2k_1}{k_1}} \\[0.3cm] f_2&=\rm 14.14 \ Hz \\[0.3cm] f_2&\approx \color{blue}{\boxed { \rm 14 \ Hz}} \end{align} {/eq}

Therefore, the frequency of oscillation of the block is {eq}\mathbf{ 14 \ Hz} {/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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