MHB Object Oscillation & Spring Shortening: Find the Answer!

AI Thread Summary
The discussion revolves around calculating how much a spring shortens when a 6.2 kg object is removed, given a spring constant (K) of 2755 N/m and an oscillation period of 0.3 seconds. Participants clarify that the relevant formula for a mass on a spring is ω = √(K/m), and they emphasize the balance between the elastic force (F_e = Ky) and gravitational force (F_G = mg) when the mass is at rest. Upon removing the mass, the spring returns to its neutral position, indicating that the displacement (y) where the mass was at rest corresponds to the spring's stretch under the weight of the object. The discussion highlights the importance of understanding the forces at play and the correct application of formulas to find the spring's shortening. The final conclusion is that the spring will return to its original length of y = 0 after the mass is removed.
leprofece
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1 object of 6.2 Kg hangs 1 balance of spring mass and performs an oscillation of each 0.3 sec If K = 2755 n/m?
How much shortens the spring by removing the object?
Use pi ^ 2 as 10
 
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leprofece said:
1 object of 6.2 Kg hangs 1 balance of spring mass and performs an oscillation of each 0.3 sec If K = 2755 n/m?
How much shortens the spring by removing the object?
Use pi ^ 2 as 10

Can you add some thoughts please?
Perhaps some relevant equations?
 
I like Serena said:
Can you add some thoughts please?
Perhaps some relevant equations?

maybe w= 2pi/t
w= 6,28/0,3

I tried (2pi)sqrt(L/g) solve to L But I don't get the 0.022 m that is the book answer

it could be get the aceleration to get the force
to try f = Kx

SO CAN anybody help me?
 
Last edited:
leprofece said:
maybe w= 2pi/t
w= 6,28/0,3

I tried (2pi)sqrt(L/g) solve to L But I don't get the 0.022 m that is the book answer

I'm afraid that is the formula for a pendulum with length L.
The formula for a mass on a spring is
$$\omega = \sqrt{\frac K m}$$

For this problem we won't need it though.
It appears you have more information than you need.
it could be get the aceleration to get the force
to try f = Kx

Good!
Let's pick $y$ for the coordinate though, to emphasize it's a vertical coordinate.
So we have an elastic force $F_e$:
$$F_e = Ky$$
And we also have the force of gravity.
$$F_G = mg$$

When the mass is at rest, the elastic force and the force of gravity have to be equal and opposite.
That is:
$$F_e = F_G$$
$$Ky = mg$$

Now if we remove the mass from the spring, the spring will return to its neutral position at $y=0$.
What can you conclude then about the $y$ where the mass was at rest before?
 
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