Question

A mass with mass 4 is attached to a spring with spring constant 25.5625 and a dashpot giving a damping 20. The mass is set in motion with initial position 1 and initial velocity 0. (All values are given in consistent units.) Find the position function x(t):

Answer 1

m = 4

k = 25.56

b = 20

`x_o=1\\v_o=0`

The differential equation for damping motion is:

`x''+\gamma x'+\omega_o^2 x=0`

where,

`\gamma = (b)/(m) = (20)/(4)=5`

`\omega_o^2=(k)/(m) = (25.56)/(4)=6.39`

Substitute the values in the differential equation and consider x" = r², x' =r and solve:

`x''+5x'+6.39x=0\\r^2+5r+6.39=0\\r=-2.5 \pm0.37i`

Therefore, solution is given by:

`x(t) = e^(-2.5t)[C_1cos0.37t+C_2sin0.37t]`
`\\`

at t = 0, x = 1

`C_1=1`

`x'(t) =-2.5e^(-2.5t)[C_1cos0.37t+C_2sin0.37t]+e^(2.5t)[-0.37C_1sin0.37t+0.37C_2cos0.37t]`

at t = 0

x' =0

`\\`
`C_2=6.76`

`x(t) = e^(-2.5t)[cos0.37t+6.76sin0.37t]`