Theory of the oscillation of a mass on a helical spring
The theory of this is really for students in the 16-19 age range but I will give it you all the same. Most textbooks for this age group have it and it is also on the web site. Don't worry if you do not follow it yet.

Consider a mass m suspended at rest from a spiral spring and let the extension produced be e. If the spring constant is k we have: mg = ke 
The mass is then pulled down a small distance x and released. 
The mass will oscillate due to both the effect of the gravitational attraction (mg) and the varying force in the spring (k(e + x)).
At any point distance x from the midpoint: restoring force = k(e + x) - mg
But F = ma, so ma = - kx and this shows that the acceleration is directly proportional to the displacement, the equation for simple harmonic motion.
The negative sign shows that the acceleration acts in the opposite direction to increasing x.

From the defining equation for simple harmonic motion. (a = - ω2 x) we have w =k/m=g/e and therefore the period of the motion T is given by: 
Period of oscillation of a helical spring (T) = 2π√(m/k).
2 5 2
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The Brainliest Answer!
T= 2 pi × sq: root of [m/k]
M- Total mass attached
K- Spring constant
T- Period of oscillation
2 5 2
good, nice
Hope it helped ^_^
Thanks for your answer
No mention :)