Simple Harmonic Motion: Period Calculation and Newton's Second Law Explanation

[SirPlus]

Homework Statement

A spring is freely hanged on a ceiling. You attach a mass to the end of the spring and let the mass go. It falls down a distance of 49 cm and comes back to where it started. It contineous to oscillate in a simple harmonic motion going up and down - a total distance of 49 cm from top to bottom. What is the period of the simple harmonic oscillation?

Homework Equations

Newtons Second Law --->> Fr = Fg

The Attempt at a Solution

To find the period of the simple harmonic motion, my first objective was to find the angular frequency using the equation (k/m) = (g/Xm) which comes from Newtons second law, however the answer I am suppose to get is ∏/$\sqrt{}10$ but i got ∏$\sqrt{}10$ ---> please explain how it works

 

[ehild]

Show your work, please. How did you get that result? What is Xm?

ehild

 

[tia89]

Given that the period of the armonic motion of the spring is given by
$$T=2\pi\sqrt{\frac{m}{k}}$$
the only thing you need to find is indeed $k$. You did correctly finding the equilibrium situation; indeed in the middle of the motion (means @$x_{1/2}=24.5$ cm) you have the gravitational force balancing the elastic one (the mass goes on because at that point it has a speed, but you do not care about this). Therefore you can find $k$ imposing
$$kx_{1/2}=mg$$
and the trick is done. Plugging numbers I indeed find $\pi/\sqrt{10}$.

 

[SirPlus]

Thanks tia89, ill provide the whole work out in the next question - I am new to the forum and I am not very good at inserting equations etc. Please understand ...