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- Thread starter leprofece
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- Mar 5, 2012

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Can you add some thoughts please?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

Perhaps some relevant equations?

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maybe w= 2pi/tCan you add some thoughts please?

Perhaps some relevant equations?

w= 6,28/0,3

I tried (2pi)sqrt(L/g) solve to L But I dont 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???

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- Mar 5, 2012

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I'm afraid that is the formula for a pendulum with length L.maybe w= 2pi/t

w= 6,28/0,3

I tried (2pi)sqrt(L/g) solve to L But I dont get the 0.022 m that is the book answer

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.

Good!it could be get the aceleration to get the force

to try f = Kx

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?