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?

Answer 1

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

$F=kx$

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

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

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

$k= \frac{(0.35)(9.8)}{12*10^{-2}}$

$k = 28.58N/m$

Perioricity in an elastic body is defined by

$T = 2\pi \sqrt{\frac{m}{k}}$

Where,

m = Mass

k = Spring constant

$T = 2\pi \sqrt{\frac{0.35}{28.58}}$

$T = 0.685s$

Therefore the period of the oscillations is 0.685s