Question

A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m. If the spring is stretched to a length of 0.5m and released with initial velocity 0, find the position of the mass at any time t. Here damping constant is zero.

Answer 1

Step-by-step explanation:

force constant of spring k = force / extension

= 35.6 / 0.5

k = 71.2 N / m

angular frequency ω of oscillation by spring mass system

`\omega = \sqrt{(k)/(m) }`

where m is mass of the body attached with spring

Putting the values

`\omega = \sqrt{(71.2)/(5) }`

ω = 3.77 radian / s

The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2

so the equation for displacement from equilibrium position that is middle point can be given as follows

x = .5 sin ( ω t + π / 2 )

= 0.5 cos ω t

= 0.5 cos 3.77 t .

x = 0.5 cos 3.77 t .