Question

An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 14.2 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:

0.04865 m

Explanation:

k = Spring Constant

m = Mass

d = Distance

g = Acceleration due to gravity = 9.81 m/s²

Angular frequency is given by

$\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{k}{m}=\omega^2\\\Rightarrow \dfrac{k}{m}=14.2^2$

At equilibrium we have

$kd=mg\\\Rightarrow d=\dfrac{mg}{k}\\\Rightarrow d=\dfrac{g}{\omega^2}\\\Rightarrow d=\dfrac{9.81}{14.2^2}\\\Rightarrow d=0.04865\ m$

The distance by which the spring stretches from its unstrained length is 0.04865 m