Question

A point mass m= 20 kg, is suspended by a massless spring of constant 2000 n/m. the point mass is released when elongation in the spring is 15 cm. the equation of displacement of particle as function of time

Answer 1

Given the spring constant we can obtain the extension of the spring when the mass is released.

According to the spring formulae :

F = kx 

Where k is the spring constant and x is the extension.

Taking g = 10 m/s²

F = ma 

F = 20 × 10 = 200N

substituting in the formula

200 = 2000x

x = 200 / 2000

x = 0.1 m

0.1 × 100 = 10 cm

The total displacement = 10 + 15 = 25 cm

This equals = 25 / 100 = 0.25 m

Taking the equation of motion :

S = ut + 1/2 at²

S is the displacement , u the initial velocity and t the time. 

The initial velocity = 0

S = 0.25 m

g = the acceleration due to gravity = 10m/s²

0.25 = 1/2 × 10 × t²

0.25 = 5t²

t² = 0.05

t = 0.22361

S = 0.22361t + 5t²

This is the expression of displacement as a function of time.