On your first trip to Planet X, you happen to take along a 195 g mass, a 40-cm-long spring, a...

Question:

On your first trip to Planet X, you happen to take along a 195 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the acceleration due to gravity on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 30.9 cm. You then pull the mass down 8.60 cm and release it. With the stopwatch, you find that 11.0 oscillations take 15.1 s. Can you now satisfy your curiosity? (In m/s{eq}^2 {/eq})

Hooke's Law:

When we join the mass to the vertical spring, the restorative force restrains the distortion of the spring and functions in the opposite direction to the force of gravity. The restorative force and the gravitational pulling force negotiate with each other and once they have attained a balance point, no subsequent vertical movement of the mass occurs. The minute the distortion, the lower the restorative force, and vice versa, so the straightforward relationship between these two parameters is as follows:

$$F \propto \Delta y $$

Or,

$$F=-k\Delta y $$

where:

  • {eq}F {/eq} is the spring force
  • {eq}k {/eq} is the spring constant
  • {eq}\Delta y {/eq} is the deformation of the spring

Answer and Explanation: 1


Given data:

  • {eq}m=\rm 195 \ g=0.195 \ kg {/eq} is the mass
  • {eq}\Delta L=\rm 30.9 \ cm=0.309 \ m {/eq} is the elongation of the spring
  • {eq}k {/eq} is the spring constant
  • {eq}T {/eq} is the period of oscillation
  • {eq}F {/eq} is the spring force


The period of oscillation is given by:

{eq}\begin{align} T&=\rm \dfrac{total \ time}{total \ number \ of \ oscillations} \\[0.3cm] &=\rm \dfrac{15.1 \ s}{11.0} \\[0.3cm] &\simeq \rm 1.37 \ s \end{align} {/eq}


The period of oscillation of the spring-mass system is:

{eq}\begin{align} T&=2\pi \sqrt{\dfrac{m}{k}} \\[0.3cm] \dfrac{m}{k}&=\dfrac{T^2}{4\pi^2} \\[0.3cm] \dfrac{k}{m}&=\dfrac{4\pi^2}{T^2} \end{align} {/eq}


Using Hooke's law, we write:

{eq}F=-k\Delta y=k\Delta L {/eq}


When the mass suspended to the spring is stationary, then using Newton's second law, we write:

{eq}\begin{align} F_{net, \ y}&=ma \\[0.3cm] F-mg&=0 \\[0.3cm] F&=mg \end{align} {/eq}


Therefore:

{eq}\begin{align} mg&=k\Delta L \\[0.3cm] g&=\left (\dfrac{k}{m} \right ) \Delta L \\[0.3cm] &=\left (\dfrac{4\pi^2}{T^2} \right ) \Delta L \\[0.3cm] &\simeq \rm \dfrac{4\pi^2}{1.37^2 \ s^2} \times 0.309 \ m \\[0.3cm] &\simeq \color{blue}{\boxed { \rm 6.50 \ m/s^2}} \end{align} {/eq}

Therefore, the acceleration due to gravity on planet X is {eq}\color{blue}{\boxed { \rm 6.50 \ m/s^2}} {/eq}.



Learn more about this topic:

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Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
202K

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.


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