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?

Respuesta :

Answer:177.26 N/m

Explanation:

Given

mass of boy [tex]m=10\ kg[/tex]

frequency [tex]f=0.97\ Hz[/tex]

For spring mass system frequency is given by

[tex]f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]

where k=spring constant

[tex]0.97=\frac{1}{2\pi }\times \sqrt{\frac{k}{10}}[/tex]

[tex]k=(2\pi \cdot 0.67)\times 10[/tex]

[tex]k=177.26\ N/m[/tex]