Question

A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

Answer 1

Here the forces are mg vertically downward and Tension of the string upwards
T= Mg
Here by applying Hook's law, the tension equals to

T = Kx [ where K is the spring constant and x is the extension under equilibrium produced in the length of the spring]

Under the condition of the equilibrium Force acting in downward direction will be equal to the force acting in upward direction

So, Tension in the string = Weight of the mass
Or, $Kx=Mg$

Or, $x=\frac{Mg}{K}$

But mass will perform SHM and it will go beyond the equilibrium position and distance traveled beyond equilibrium position will be $\frac{Mg}{K}$

Hence, maximum extension produced in the string is $\frac{2Mg}{K}$