Question

A 2.00 kg object is attached to a spring and placed on frictionless, horizontal surface. Ahorizontal force of 18.0 N is required to hold the object at rest when it is pulled 0.250 mfrom its equilibrium position. What is the angular frequency, in rad/s, of the oscillation?

Answer 1

Answer:

6 rad/s

Explanation:

In a spring the angular frequency is calculated as follows:

$\omega=\sqrt{\frac{k}{m} }$

where $\omega$ is the angular frequency, $m$ is the mass of the object in this case $m=2kg$, and $k$ is the constant of the spring.

To calculate the angular frequency, first we need to find the constant $k$ which is calculated as follows:

$k=\frac{F}{x}$

Where $F$ is the force: $F=18N$, and $x$ is the distance from the equilibrium position: $x=0.25m$.

Thus the spring constant:

$k=\frac{18N}{0.25m}$

$k=72N/m$

And now we do have everything necessary to calculate the angular frequency:

$\omega=\sqrt{\frac{k}{m} }=\sqrt{\frac{72N/m}{2kg} }=\sqrt{36}$

$\omega=6rad/s$

the angular frequency of the oscillation is 6 rad/s