Simple harmonic frequency clarification

  • Thread starter Rasine
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  • #1
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A massless spring with spring constant 23.5 N/m hangs vertically. A body of mass 0.30 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released.

What is the frequency of oscillation of the resulting motion, assumed to be simple harmonic?

ok so f=(1/2pi)[sqroot(k/m)] cuz w=sqroot(k/m) and f=w/2pi

so i have (1/2pi)[sqroot(23.5/.30)] which is 13.90.....right?

and the units would be Hz right becuase w is in rad/s and that divided by rad would give cycles/s


i think i am doing something wrong. please point it out to me.
 

Answers and Replies

  • #2
AlephZero
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so i have (1/2pi)[sqroot(23.5/.30)] which is 13.90.....right?
Your thinking and your equation are OK. Maybe your calculator is broken :smile:

Hint: my calculator says sqrt(23.5/0.3) = 8.85
 
  • #3
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thats what i get too but the formula says that 8.85 should be divided by 2pi which is 13.90

is that calculation right? should i do that?
 
  • #4
AlephZero
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On my calculator 8.85 x PI / 2 = 13.90. I guess you are pushing the wrong buttons.

2pi is about 6.28. There is no way that 8.85/6.28 could be 13.90 !!
 

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