Block hanging from spring when a second block is added

[dtesselstrom]

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.

 

[OlderDan]

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?

 

[kyle8921]

Based on the given information, it seems like the second block causes the original block to sag by 3.00 cm, indicating that the added weight has affected the equilibrium position of the system. This means that the spring constant (k) of the spring has also changed.

To accurately calculate the oscillation frequency of the system, we need to take into account the new equilibrium position and the new spring constant. The correct equation to use would be T=2pi sq root of m/(k+2k), where m is the mass of one block and k is the original spring constant.

In order to solve for k, we can use the given information that the original block sags by 3.00 cm. This means that the new equilibrium position of the system is 3.00 cm below the original equilibrium position. We can use this to set up the equation kx=mg, where x is the displacement from the original equilibrium position and m is the mass of one block.

Solving for k, we get k=mg/x. Plugging this into the equation for T, we get T=2pi sq root of m/((mg/x)+2(mg/x)) = 2pi sq root of m/(3mg/x) = 2pi sq root of 3x/g.

Therefore, the oscillation frequency of the system would be 2pi sq root of 3x/g, where x is the displacement from the original equilibrium position (in this case, 3.00 cm).

I hope this helps you in solving the problem. It is important to carefully consider all the given information and use the appropriate equations to accurately calculate the desired result.