Question

When a 2-kg block is suspended from a spring, the spring is stretched a distance of 20 mm. Determine the natural frequency and the period of vibration for a 0.5-kg block attached to the same spring

Answer 1

Answer:

7.05 Hz

Explanation:

The natural frequency of a mass-spring system is:

$f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}$

To determine the constant k of the spring we use Hooke's law:

Δl  = F / k

k = F / Δl

In the first case the force was the weight of the 20 kg mass and the Δl was 20 mm.

F = m * a

F = 2 * 9.81 = 19.6 N

Then:

k = 19.6 / 0.02 = 980 N/m

Therefore:$f = \frac{1}{2 \pi}\sqrt{\frac{980}{0.5}} = 7.05 Hz$