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) 0.286 s
B) 1.44 s
C) 0.695 s
D) 0.0769 s
E) 0.483 s

Question

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)  0.286 s
B)  1.44 s
C)  0.695 s
D)  0.0769 s
E)  0.483 s

Quizplus · Accepted Answer

The period of oscillation 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). Thus, $k = \frac{mg}{x} = \frac{0.350 \times 9.8}{0.12}$. Then, substituting $m$ and $k$ into the formula for $T$, we solve for the period.

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