Block hanging from spring when a second block is added

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A block hanging from a vertical spring experiences a sag of 3.00 cm when a second identical block is added, indicating a change in equilibrium. The user initially attempted to calculate the spring constant (k) using the formula F = -ky, where F is the weight of the two blocks and y is the sag. However, the calculated oscillation frequency using T = 2π√(m/k) was incorrect. The discussion emphasizes the need to accurately determine how much the spring stretches from its natural length with the additional weight. Clarifying the relationship between the sag and the spring's properties is essential for solving the problem correctly.
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A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by 3.00 cm. It wants to know the oscillation frequency of this problem.
So far I tried solving for F=-ky with F being 2m*g and y equal to .03. I solved for k and plugged that into T=2pi sq root of m/k this however said it was wrong. Any information on how to approach this would be nice.
 
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dtesselstrom said:
A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by 3.00 cm. It wants to know the oscillation frequency of this problem.
So far I tried solving for F=-ky with F being 2m*g and y equal to .03. I solved for k and plugged that into T=2pi sq root of m/k this however said it was wrong. Any information on how to approach this would be nice.
The 3cm sag comes from adding 1m. How much do you think the spring is streteched from its natural length with the 2m hanging on it?
 

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