Answer:
The distance by which the spring stretches is 1.48 cm.
Explanation:
Given that,
An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s, [tex]\omega=25.7\ rad/s[/tex]
We know that angular frequency in SHM is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\\omega^2=\dfrac{k}{m}\\\\\dfrac{k}{m}=(25.7)^2.............(1)[/tex]
When the object is allowed to hang stationary from it, the force due to spring is balanced by its weight. Such that :
[tex]kd=mg\\\\d=\dfrac{g}{(k/m)}[/tex]
From equation (1) :
[tex]d=\dfrac{9.8}{(25.7)^2}\\\\d=0.0148\ m\\\\d=1.48\ cm[/tex]
So, the distance by which the spring stretches is 1.48 cm.
