How to Calculate the Spring Constant for Simple Harmonic Motion?

AI Thread Summary
To calculate the spring constant for a vertical spring in simple harmonic motion, the initial problem involves a 2kg weight that stretches the spring by 0.25m. The spring constant (k) can be determined using Hooke's law, where k is derived from the force exerted by the weight (mg) and the displacement (x). The correct period (T) for 10 oscillations is calculated using the formula T = 2π√(m/k), leading to a discrepancy in the expected value of k. The discussion highlights the importance of applying the second law of dynamics (F=ma) and the conservation of mechanical energy to find k accurately. Ultimately, the correct value of k should align with the observed oscillation period of 10 seconds.
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simple harmonic motion ?

hey guys, I am having troube with this problem

a vertical spring stretches .25m when a 2kg weight is hung from it. If that weight were oscillating up and down, how long would it take to complete 10 cycles?

so first i tried to find k by using PEi + KEi + PE(spring)i =PEf etc

so 2kg(.25m)(9.8) = .5(k)(.25)^2

so k= 156.8

then finding T T= 2(pi) sqroot(m/k)

so i get T= .709 * 10 oscillations = 7.1sec

correct answer is 10sec, meaning T must equal 1sec, which means k should equal somewhere around 78 or 79 ...working backwards...
i don't see how they get this value...any help is greatly appriciated THANKS
 
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Hmm. Why can't you get the "k" value from the second law of dynamics applied to the spring-mass system...?

Daniel.
 
hahaha, right on man, thanks! also thanks for not making me look like an idiot, i totally spaced using F=ma,,,,,however my question still remains, is it possible using MEi = MEf to obtain spring constant? thanks again for the quick help...
 
physics noob said:
a vertical spring stretches .25m when a 2kg weight is hung from it.
From that fact and Hooke's law you should be able to calculate the spring constant.
 
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