A 5-kg block is attached to a horizontal spring of constant 500 N/m.It is pulled 20 cm from its equilibrium position and released.What is its angular frequency?
A)12.2 rad/s
B)10.0 rad/s
C)15.5 rad/s
D)17.6 rad/s
E)11.2 rad/s

Question

Quizplus · Accepted Answer

The angular frequency, $\omega$, of a mass-spring system is given by $\omega = \sqrt{\frac{k}{m}}$, where $k$ is the spring constant and $m$ is the mass. Plugging in the given values, $k = 500\, \text{N/m}$ and $m = 5\, \text{kg}$, we get $\omega = \sqrt{\frac{500}{5}} = \sqrt{100} = 10\, \text{rad/s}$.

A 5-Kg Block Is Attached to a Horizontal Spring of Constant