Motion of an object attached to a spring

[myoplex11]

Homework Statement

A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately?

Homework Equations

X(t) = Acos(wt+φ)


3. The Attempt at a Solution
the awnser to this problem is 0.05cos10t. I see where they get the 0.05 from since it is the amplitude but how do they figure out w(angular frequency)=10

 

[tyco05]

w = sqrt (k/m)

F = -kx

Now you have the equations you need

 

[myoplex11]

how do i figure out F acceleration is not given

 

[tyco05]

F = -kx

is Hooke's Law.

The force comes from the weight of the body. And x is the extension of a spring.


Use this to find k.

Then apply the other formula to find w.