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:177.26 N/m

Explanation:

Given

mass of boy $m=10\ kg$

frequency $f=0.97\ Hz$

For spring mass system frequency is given by

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

where k=spring constant

$0.97=\frac{1}{2\pi }\times \sqrt{\frac{k}{10}}$

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

$k=177.26\ N/m$