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anonymous
 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.
anonymous
 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is w= the square root of k/m

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hi,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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0At 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@igbona,if you apply conservation of energy in such a problem then you can easily get your answer. if any query ,tell me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so, the x in (1/2)Kx^2 can be substituted by A => (1/2)KA^2. why?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea hence you use that formula in amplitude point this point has x=A did you got it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because of at extreme position the displacement of the block will be X=A,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ igbona,any problem,,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Taufique, none at all. I was just curious. Thank you for your help.
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