The spring constant of the spring on which the block is attached is 7,500 N/m.
The given parameters;
- mass of the block, m = 3 kg
- simple harmonic equation, x = 2cos(50t)
The angular frequency is calculated as;
ωt = 50t
ω = 50 rad/s
The spring constant of the spring is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = \omega ^2 m\\\\k = (50)^2 (3)\\\\k = 7,500 \ N/m[/tex]
Thus, the spring constant of the spring is 7,500 N/m.
Learn more here:https://brainly.com/question/22499689
