When a 0.350-kg package is attached to a vertical spring and lowered slowly,the spring stretches 12.0 cm.The package is now displaced from its equilibrium position and undergoes simple harmonic oscillations when released.What is the period of the oscillations?
A)0.695 s
B)0.483 s
C)0.286 s
D)0.0769 s
E)1.44 s

Question

Quizplus · Accepted Answer

The period of oscillations for a mass-spring system is given by $T = 2\pi\sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant. First, we find $k$ using $F = kx$, where $F = mg$ and $x$ is the displacement (0.12 m). With $g = 9.8 \, \text{m/s}^2$, $k = \frac{mg}{x} = \frac{0.350 \times 9.8}{0.12}$. Then, plug $m$ and $k$ into the formula for $T$.

When a 0350-Kg Package Is Attached to a Vertical Spring and Lowered