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

Step-by-step explanation:

Given

mass of boy
`m=10\ kg`

frequency
`f=0.97\ Hz`

For spring mass system frequency is given by

`f=(1)/(2\pi )\sqrt{(k)/(m)}`

where k=spring constant

`0.97=(1)/(2\pi )* \sqrt{(k)/(10)}`

`k=(2\pi \cdot 0.67)* 10`

`k=177.26\ N/m`