Question

An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s. Calculate the distance d by which the spring stretches from its unstrained length when the object is allowed to hang stationary from it.

Answer 1

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, $\omega=25.7\ rad/s$

We know that angular frequency in SHM is given by :

$\omega=\sqrt{\dfrac{k}{m}} \\\\\omega^2=\dfrac{k}{m}\\\\\dfrac{k}{m}=(25.7)^2.............(1)$

When the object is allowed to hang stationary from it, the force due to spring is balanced by its weight. Such that :

$kd=mg\\\\d=\dfrac{g}{(k/m)}$

From equation (1) :

$d=\dfrac{9.8}{(25.7)^2}\\\\d=0.0148\ m\\\\d=1.48\ cm$

So, the distance by which the spring stretches is 1.48 cm.