Solving Spring Oscillations Homework: Stretched Distance d

AI Thread Summary
The problem involves a 0.35 kg mass attached to a vertical spring, which stretches a distance d when lowered. The mass undergoes 100 oscillations in 48.9 seconds, leading to a calculated period of approximately 0.489 seconds. The spring constant k was determined to be 3.45 N/m using the formula T = 2π(√(m/k)). The gravitational force mg was calculated as 3.43 N, and the resulting distance x was found to be around 1, but the expected answer is 5.94 cm. The discrepancy suggests a miscalculation in the application of the formulas or assumptions made during the solution process.
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Homework Statement


When a mass of 0.35 kg is attached to a vertical spring and lowered slowly, the spring stretches a distance of d. The mass is now displaced from its equilibrium position and undergoes 100 oscillations in 48.9 seconds. WHat is the stretched distance d?

Homework Equations


Number of oscillations/Time = Period (T)
T = 2pi (\sqrt{m/k})
F=kx

The Attempt at a Solution


I found by doing 100/48.9, the period is around 2 oscillations per second.
Using the second equation I figured out that k=3.45 by plugging it all the other variables. Then i know that the F=mg. So mg is 3.43.
3.43/3.45 = x (distance)
The distance comes out to around 1. But my teacher says that 5.94 cm...what did i do wrong?
 
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the period is supposed to be 48.9/100 s
 

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