Question

A mass-spring-damper system has a mass of 100 kg. Its free response amplitude decays such that the amplitude of the 30th cycle is 20% of the amplitude of the 1st cycle. It takes 60 s to complete 30 cycles. Estimate the damping constant c and the spring constant k.

Answer 1

Answer:

the value for the damping constant is c = 5.3408Ns/m and the spring constant is k is 987N/m

Explanation:

Consider the formula for the logarithmic decrement (δ)

δ = $\frac{1}{n} In(\frac{B_1}{B_{n+1}} )----(1)$

Consider the formula for damping ratio  (C)

C =δ / √4π² + ² ----------(2)

The given amplitude of the 30th cycle is 20% amplitude of the 1st cycle

The ratio of last peak to first peak is

$\frac{B_{n+1}}{B_1} =20$%--------------(3)

where

n is the number of cycle

Re arrange eqn (3)

$\frac{B_{n+1}}{B_1} = \frac{100}{5} \\\\\frac{B_{n+1}}{B_1} = 5-----(4)$

The given value for n is 30

substitute 30 for n and eqn (4) in eqn(1)

δ = $\frac{1}{30} In5$

   $= 0.0536$

substitute  0.0536 for δ in eqn (2)

$C = \frac{0.0536}{\sqrt{39.4784+0.00287} } \\C = 0.0085----(5)$

Consider the formula for damping ratio  (C)

C = $\frac{c}{2\sqrt{mk} } ----(6)$

where

c is the damping constant

m is the mass

substitute 100 for m  in eqn(6)

$C = \frac{c}{2\sqrt{100k} } \\\\C = \frac{c}{20\sqrt{k} } ----(7)$

compute eqn(5) and eqn(7)

$\frac{c}{20\sqrt{k} } = 0.0085\\\\c = 0.0085(20\sqrt{k} )\\\\c= 0.17\sqrt{k} ----(8)$

The given time to complete 30 cycles in 60s is

period = 60 / 30

= 2s

Formula for spring constant

$k =mw^2_n----(9)$

where

$w_n$ is the undamped natural frequency

consider the formula for damped natural frequency

$w_d=w_n\sqrt{1-C^2}$

rearrange

$w_n = \frac{w_d}{\sqrt{1-C^2} } ----(10)$

substitute eqn (10) to eqn (9)

$k = m(\frac{w_d}{\sqrt{1-C^2} } )\\\\k=\frac{mw^2_d}{1-C^2} ---(11)$

consider the formula for the damped natural frequency

$w_d=\frac{2\pi }{P} ----(12)$

consider eqn (12) in eqn (11)

$k = \frac{m(\frac{2\pi }{P} )^2}{1-C^2}$

substitute 100 for m, 2 for P, 0.0085 for C in the eqn above

$k = \frac{(100)(\frac{2\pi }{2})^2 }{1-(0.0085)^2} \\\\k= \frac{986.96}{0.9999} \\\\k=987N/m$

substitute 987 for k in eqn (8)

$c=0.17\sqrt{987} \\c=0.17(31.4166)\\c=5.3408Ns/m$

Thus, the value for the damping constant is c = 5.3408Ns/m and the spring constant is k is 987N/m