Question

A spring executes SHM with mass of 4 kg attached to it. The force constant of spring is 100 N/m. If at any instant its velocity is 40 cm/s, the displacement from the mean position will be (Here amplitude is 0.5 m)​

Answer 1

Answer:

please explain the question

Answer 2

Explanation:

Given A spring executes SHM with mass of 4 kg attached to it. The force constant of spring is 100 N/m. If at any instant its velocity is 40 cm/s, the displacement from the mean position will be  

• Now we need to find the displacement at the point. So given
• Mass m = 4kg, amplitude A = 0.5 m, velocity v1 = 40 cm / s = 0.4 m / s, force constant k = 100 N/m
• So displacement x1 = ?
• Now the relation between displacement and velocity is given as
• So v^2 = ω^2 (A^2 – x^2)
• So ω^2 = k / m
•             = 100 / 4
• Or ω^2 = 25
• Therefore v1^2 =  25 (0.25 – x1^2)
•               0.16 = 6.25 – 25 x1^2
•            So – 6.09 = - 25 x1^2
•            So 25 x1^2 = 6.09
•         Or x1^2 = 0.2436
•        Or x1 = 0.49 or 0.5 m    

Reference link will be

https://brainly.in/question/8860744