What is the stiffness of the spring in a child's pogo stick?

  • Thread starter Charles Stark
  • Start date
In summary, the conversation discusses finding the stiffness of a spring in a child's pogo stick by using the equation T = 2π√(m/k). The student first attempts to solve for k by using the frequency/period relationship, but is reminded to keep track of units. The student then reworks the problem and finds the correct answer of 350 N/m, which is the standard unit for stiffness. The conversation also touches on the importance of significant figures in calculations.
  • #1
Charles Stark
31
5

Homework Statement


Find the stiffness of the spring in a child's pogo stick if the child has a mass of 35kg and bounces once every 2.0 seconds.

T = 2.0 seconds m = 35kg

Homework Equations


T = 2π√(m/k)

The Attempt at a Solution


Using the above, I figured k = mπ2 which worked out to be 345.44

This problem didn't have an answer in the back of the book. I saw a similar equation for frequency so I used the frequency/period relationship to formulate the above equation. I figured I should ask for a check in case I assumed wrong.
 
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  • #2
You have some minor issues with your k=m pi squared form. The units don't work because you did some cancellation and you were somewhat sloppy. What are the units for k? What are the units for m?

When you do problems like this be sure to keep your steps and show them. And in particular, be sure that you check that the units work.
 
  • #3
DEvens said:
You have some minor issues with your k=m pi squared form. The units don't work because you did some cancellation and you were somewhat sloppy. What are the units for k? What are the units for m?

When you do problems like this be sure to keep your steps and show them. And in particular, be sure that you check that the units work.
The units do work.
T = 2π√(m/k)
substituting value of T as 2s,
2 = 2π√(m/k)
1 = π√(m/k)
√(k/m) = π
now, squaring on both the sides,
k/m = π^2
k =mπ^2
 
  • #4
Read my post again please. What are the units of k? What are the units of m? Just because 2 seconds divided by 2 gives you 1 *something* does not mean the units work when you discard that *something*.
 
  • #5
DEvens said:
You have some minor issues with your k=m pi squared form. The units don't work because you did some cancellation and you were somewhat sloppy. What are the units for k? What are the units for m?

When you do problems like this be sure to keep your steps and show them. And in particular, be sure that you check that the units work.

I'll re-work it again and see if I can find a different way
 
  • #6
Charles Stark said:
I'll re-work it again and see if I can find a different way
Your method is ok. I think all DEvens is complaining about is that you did not keep track of units, so didn't state the units in the answer. But with the most obvious filling in of those omissions, I believe your answer is numerically correct.
 
  • #7
I actually found the answer to it. It states 350 N/m which I can see from rounding but it didn't say anything about estimating in the problem. Maybe an unfortunate mistake
 
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  • #8
Charles Stark said:
I actually found the answer to it. It states 350 N which I can see from rounding but it didn't say anything about estimating in the problem. Maybe an unfortunate mistake
In principle, you should infer the number of significant figures to quote in the answer from (the fewest of) those provided. The clue here is 2.0 seconds, not 2.00 or 2.
 
  • #9
haruspex said:
In principle, you should infer the number of significant figures to quote in the answer from (the fewest of) those provided. The clue here is 2.0 seconds, not 2.00 or 2.

This would make sense since I always forget about significant figures.
 
  • #10
the units of k is NOT Newtons! How IS Force related to the spring stiffness?
 
  • #11
lightgrav said:
the units of k is NOT Newtons! How IS Force related to the spring stiffness?

Oops that was supposed to be 350 N/m not N.
 
  • #12
Every spring's stiffness is described by k N/m ... that's how we know it is a stiffness - way more important than the number!
 

FAQ: What is the stiffness of the spring in a child's pogo stick?

What is Hooke's Law?

Hooke's Law is a principle in physics that states the force needed to extend or compress a spring is directly proportional to the distance the spring is stretched or compressed.

Who discovered Hooke's Law?

Robert Hooke, an English scientist, first proposed the law in the 17th century.

What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the distance the spring is stretched or compressed.

How does Hooke's Law relate to oscillation?

Hooke's Law is the basis for understanding the behavior of oscillating systems, such as springs and pendulums. The force of the spring is what allows the object to oscillate back and forth.

What are some real-life applications of Hooke's Law?

Hooke's Law has many practical applications, including in the design of shock absorbers, car suspensions, and other mechanical systems. It is also used in the study of earthquakes and other natural phenomena.

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