A mass of 445 grams is suspended by a spring. The spring stretches a distance of 5.60 cm under...

Question:

A mass of 445 grams is suspended by a spring. The spring stretches a distance of 5.60 cm under the load of the mass. What is the spring constant? Suppose that 125 g falls off the spring. What would be the period of the resulting oscillation?

Spring:


If a spring is at the unstretched position and a mass (m) is vertically attached to it. The attached mass will stretch the sping in the downward direction and the magnitude of stretch is directly proportional to the weight of the attached mass.

Answer and Explanation: 1

We are given the following data:

  • Mass suspended by the spring, {eq}m=445\ \text{g} {/eq}
  • Displacement of spring, {eq}x=5.60\ \text{cm} {/eq}


If {eq}(F) {/eq} is the external force applied to the spring to stretch or compress it by a distance {eq}(x) {/eq} from its unstretched or initial position, we can compute the force by using Hooke's law which is expressed as follow:

{eq}\begin{align} F&=kx\\[0.3 cm] k&=\boxed{\dfrac{F}{x}} \end{align} {/eq}

Here the external force is the weight attached to the spring.

Substituting values in the above equation, we have:

{eq}\begin{align} k&=\dfrac{F}{x}\\[0.3 cm] &=\dfrac{mg}{x}\\[0.3 cm] &=\dfrac{445\times10^{-3}\ \text{g}\times9.80\ \text{m/s}^{2}}{5.60\times10^{-2}\ \text{m}}\\[0.3 cm] &=77.875\ \text{N/m}\\[0.3 cm] &\approx\boxed{\color{red}{77.9\ \text{N/m}}} \end{align} {/eq}


The time period of one complete cycle of vibration is expressed by the following equation:

{eq}\boxed{T=2\pi\sqrt{\dfrac{m'}{k}}} {/eq}

Where

  • m' is the mass attached to the spring
  • k is the force constant of the spring


Now if 125 g of mass falls off the spring, the magnitude of the remaining mass attached to the spring is:

{eq}\begin{align} m'&=445\ \text{g}-125\ \text{g}\\[0.3 cm] &=320\ \text{g} \end{align} {/eq}


Substituting values in the above equation, we have:

{eq}\begin{align} T&=2\pi\sqrt{\dfrac{m'}{k}}\\[0.3 cm] &=2\pi\sqrt{\dfrac{320\times10^{-3}\ \text{kg}}{77.875\ \text{N/m}}}\\[0.3 cm] &=\boxed{\color{red}{0.402\ \text{s}}} \end{align} {/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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