Mass-Spring Oscillation question

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The discussion centers on calculating the oscillation frequency of a two-block system suspended from a vertical spring. The original poster initially calculated the period as 0.40 seconds, leading to a frequency of 2.5 Hz, which was incorrect. The correct approach involves determining the spring constant (k) using the equation mg = ky, where m is the mass of one block, g is the acceleration due to gravity, and y is the displacement of 8.00 cm. The correct mass for the oscillation period formula T = 2π√(m/k) should account for both blocks in the system.

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Mozart
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I tried to work out this problem a few different ways but I never get the right answer.

A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by 8.00 cm

What is the oscillation frequency of the two-block system?

What I've done so far:

Attempt 1:

(Fnet)y=-ky
2mg=-ky
m=((-ky)/(2g))

Then using T=2pi times sqaure root[m/k]

puting in my m as what I found in terms of k, y, and g. The K's cancel and I am left with things I know and then calculate to find my T it turns out to be 0.40 seconds

and then I finish off my problem with freqency= 1/period and get 2.5 Hz

But this is incorrect.

I probably made a wrong assumption but I can't put my finger on it. I hope someone can put me on the right track.

Thank you.
 
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Mozart said:
What I've done so far:

Attempt 1:

(Fnet)y=-ky
2mg=-ky
m=((-ky)/(2g))
What you want to do is find the spring constant in terms of m, y, & g. Assuming that y is the additional displacement from equilibrium, then:
mg = ky (since one block is added)

Then using T=2pi times sqaure root[m/k]
What mass goes here? (The system has two blocks now.)
 
Thank you! That worked perfectly. I understand where I made my mistake now too.
 

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