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igbona

  • 4 years ago

A simple harmonic oscillator with m= 0.560kg and total energy E = 130J has amplitude A= 1.65m . Find the spring constant.

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  1. igbona
    • 4 years ago
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    in order to find k i think we use \[w=\sqrt{k/m}\] but we need to find w to find k. I don't know if this is the right way to solve the problem.

  2. igbona
    • 4 years ago
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    it is w= the square root of k/m

  3. hosein
    • 4 years ago
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    Hi,i think when mass pass origin all of it's E is EK so\[E=(1/2)mv _{2}^{\max}\]&\[v _{\max}=A \omega\]so you can find omeg asd from that you can yield to k

  4. Taufique
    • 4 years ago
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    At the extreme position ,the block has Zero velocity. Total mechanical energy=K.E+P.E=>130=0+1/2KA^2=>K=(260/A^2)=260/(1.65*1.65)=95.5Nm...

  5. Taufique
    • 4 years ago
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    @igbona,if you apply conservation of energy in such a problem then you can easily get your answer. if any query ,tell me.

  6. igbona
    • 4 years ago
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    so, the x in (1/2)Kx^2 can be substituted by A => (1/2)KA^2. why?

  7. hosein
    • 4 years ago
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    yea hence you use that formula in amplitude point this point has x=A did you got it?

  8. igbona
    • 4 years ago
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    yes :)

  9. Taufique
    • 4 years ago
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    because of at extreme position the displacement of the block will be X=A,

  10. Taufique
    • 4 years ago
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    @ igbona,any problem,,

  11. igbona
    • 4 years ago
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    @Taufique, none at all. I was just curious. Thank you for your help.

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