SHM Find amplitude from ω and x

[dahoom102]

Homework Statement

A spring with a natural length 40 cm and a spring constant of 400 N/m is hung vertically with a 10 kg mass attached to the end. Assuming the spring's mass is negligible, what will be the final length of the spring when it reaches equilibrium?

Relevant Equations

x=A*cos(ωt)

x=A at equilibrium

ω=sqrt(k/m)

Hi,

I have no idea what formula to use while given these values, basically, it fits no formula. Any thing could help?

Many thanks in advance

Correct answer is 65

 

[PeroK]

What is the relevance of SHM to this problem?

 

[dahoom102]

What is the relevance of SHM to this problem?

It came in a simple harmonic motion quiz so it should be relevant i guess also its about a spring and amplitude

 

[PeroK]

It came in a simple harmonic motion quiz so it should be relevant i guess also its about a spring and amplitude

What is "equilibrium" in relation to SHM?

Assuming the spring's mass is negligible, what will be the final length of the spring when it reaches equilibrium?

 

[dahoom102]

What is "equilibrium" in relation to SHM?

It is the point at which a body exhibiting SHM oscillates between

 

[PeroK]

It is the point at which a body exhibiting SHM oscillates between

Why not calculate that then?

 

[dahoom102]

Why not calculate that then?

Ok but i am still unable to grasp how is this going to help

mg=kx

x=98/400=0.245m

 

[PeroK]

Ok but i am still unable to grasp how is this going to help

mg=kx

x=98/400=0.245m

That looks about right.

 

[PeroK]

Let's call it $25cm$. What's the difference between $25cm$ and the $65cm$ that the answer is supposed to be?

 

[dahoom102]

Let's call it $25cm$. What's the difference between $25cm$ and the $65cm$ that the answer is supposed to be?

Ohhhhh i see now! Thanks a ton PeroK for your time!