Answer:
Spring constant, k = 5483.11 N/m
Explanation:
It is given that,
Mass of the organ, m = 2 kg
The natural period of oscillation is, T = 0.12 s
Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
[tex]k=\dfrac{4\pi^2 m}{T^2}[/tex]
[tex]k=\dfrac{4\pi^2 \times 2\ kg}{(0.12\ s)^2}[/tex]
k = 5483.11 N/m
So, the spring constant for the spring in the scientist's model is 5483.11 N/m.
