# Mass on spring

#### leprofece

##### Member
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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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?

#### leprofece

##### Member
Perhaps some relevant equations?
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

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

SO CAN anybody help me???

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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
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 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?