Question

1. A popular baby toy is called a “Johnny Jump Up.” The Johnny Jump Up is a seat that a 12 month old would sit it. The seat is attached to a long spring that is connected to the frame of a door. If a 10kg baby bobs up and down with a frequency of 0.67 Hz, what is the spring constant of the spring in the Johnny Jump Up?

Answer 1

Answer:177.2 N/m

Explanation:

Given

Frequency of oscillation $f_o=0.67\ Hz$

Mass of boy $m_o=10\ kg$

Spring mass system is given by

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

$f_o=\frac{1}{2\pi }\times \sqrt{\frac{k}{m_o}}$

where k=Spring constant

$k=(2\pi \cdot f_o)^2\times m_o$

$k=(2\pi \times 0.67)^2\times 10$

$k=(4.21)^2\times 10$

$k=17.72\times 10$

$k=177.2\ N/m$