Question

A vertical spring stretches 3.4 cm when a 12-g object is hung from it. The object is replaced with a block of mass 26 g that oscillates up and down in simple harmonic motion. Calculate the period of motion.

Answer 1

Answer:

Period of motion is approximately 0.5447  seconds

Explanation:

We start by calculating the constant "k" of the spring which can be derived from the fact that an object of mass 12 g produced a stretch of 3.4 cm: (we write everything in SI units)

F = k * x

0.012 kg * 9.8  m/s^2 = k 0.034 m

k = 0.012 kg * 9.8  m/s^2 / (0.034 m)

k = 3.46 N/m

now we use the formula for the period (T) of a spring of constant k with a hanging mass 'm':

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

which in our case becomes:

$T=2\pi\,\sqrt{\frac{0.026}{3.46} } \approx 0.5447\,\,sec$