A 500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. If the total energy of the system is 4J, find the force constant of the spring?

Respuesta :

Answer:

315.5 N/m

Explanation:

m = 500 g = 0.5 kg

T = 0.25 second

Total energy, E = 4 J

Let K be the spring constant.

The formula for the time period is given by

[tex]T = 2\pi \sqrt{\frac{m}{K}}[/tex]

[tex]0.25 = 2\times 3.14\times \sqrt{\frac{0.5}{K}}[/tex]

[tex]0.0398=\sqrt{\frac{0.5}{K}}[/tex]

[tex]1.585\times 10^{-3}={\frac{0.5}{K}}[/tex]

K = 315.5 N/m