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igbona
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A simple harmonic oscillator with m= 0.560kg and total energy E = 130J has amplitude A= 1.65m . Find the spring constant.
 2 years ago
 2 years ago
igbona Group Title
A simple harmonic oscillator with m= 0.560kg and total energy E = 130J has amplitude A= 1.65m . Find the spring constant.
 2 years ago
 2 years ago

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igbona Group TitleBest ResponseYou've already chosen the best response.0
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 years ago

igbona Group TitleBest ResponseYou've already chosen the best response.0
it is w= the square root of k/m
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Taufique Group TitleBest ResponseYou've already chosen the best response.2
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...
 2 years ago

Taufique Group TitleBest ResponseYou've already chosen the best response.2
@igbona,if you apply conservation of energy in such a problem then you can easily get your answer. if any query ,tell me.
 2 years ago

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

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

Taufique Group TitleBest ResponseYou've already chosen the best response.2
because of at extreme position the displacement of the block will be X=A,
 2 years ago

Taufique Group TitleBest ResponseYou've already chosen the best response.2
@ igbona,any problem,,
 2 years ago

igbona Group TitleBest ResponseYou've already chosen the best response.0
@Taufique, none at all. I was just curious. Thank you for your help.
 2 years ago
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