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Spring constant of a thread

  • #1
138
0

Homework Statement



A 2.60 g spider is dangling at the end of a silk thread. You can make the spider bounce up and down on the thread by tapping lightly on his feet with a pencil. You soon discover that you can give the spider the largest amplitude on his little bungee cord if you tap exactly once every 4.90 s seconds. What is the spring constant of the silk thread.

Homework Equations



[tex]\omega[/tex]= 2[tex]\pi[/tex]/T
[tex]\omega[/tex]= [tex]\sqrt{}k/m[/tex]

The Attempt at a Solution


so i equated the 2 equations and tried to isolate for k and this is what my equation looked like : m*(2[tex]\pi[/tex]/T)^2 = k ..i tried rearranging twice and my answer and equation are coming out the same ..pls. help
 

Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
so i equated the 2 equations and tried to isolate for k and this is what my equation looked like : m*(2[tex]\pi[/tex]/T)^2 = k ..i tried rearranging twice and my answer and equation are coming out the same ..pls. help
That looks fine to me. Why do you think it's not?
 
  • #3
138
0
when i put the answer into my hwk the answer is appparently wrong..i was also wondering if i was right in assuming the period to be 4.90 seconds??
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,572
192

Homework Statement



A 2.60 g spider is dangling at the end of a silk thread. You can make the spider bounce up and down on the thread by tapping lightly on his feet with a pencil. You soon discover that you can give the spider the largest amplitude on his little bungee cord if you tap exactly once every 4.90 s seconds. What is the spring constant of the silk thread.

Homework Equations



[tex]\omega[/tex]= 2[tex]\pi[/tex]/T
[tex]\omega[/tex]= [tex]\sqrt{}k/m[/tex]

The Attempt at a Solution


so i equated the 2 equations and tried to isolate for k and this is what my equation looked like : m*(2[tex]\pi[/tex]/T)^2 = k ..i tried rearranging twice and my answer and equation are coming out the same ..pls. help

What value did you get ? Did you convert the mass to kg ? (I assume they want the k in N/m)
 
  • #5
138
0
the value i am getting is 4.85x10^-3 N/m
 
  • #6
Doc Al
Mentor
44,882
1,129
i was also wondering if i was right in assuming the period to be 4.90 seconds??
Yes, that was correct.
 
  • #7
Doc Al
Mentor
44,882
1,129
the value i am getting is 4.85x10^-3 N/m
Double check your arithmetic.
 
  • #8
138
0
thankx i got it :))
 

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