Question

A 0.5 kg block is attached to a spring (k = 12.5 N/m). The damped frequency is 0.2% lower than the natural frequency, (a) What is the damping constant? (b) How does the amplitude vary with time? (c) Determine the critical damping constant?

Answer 1

Answer:

a)  1.58 kg s^{-1}

b)  x_m e^{-1.58t}   x_m is initial amplitude

c) 5 kg s^{-1}

Explanation:

given data:

mass =0.5 kg

k = 12.5 N/m

from the data given

a) $w_d = w_o - \frac{0.2}{100}w_o$

$= w_o - 0.002w_o = 0.99w_o$

$w_d = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}$

$0.998w_o = \sqrt{w_o^2 - \frac{b^2}{4m^2}$

$(0.998w_o)^2 = w_o^2 -\frac{b^2}{1}$

$b^2 = w_o^2 -(0.998w_o)^2$

$b^2 = w_o^2(1-0.998^2) = 3.996 *10^{-3} w_o^2$

$b = w_o\sqrt{3.996*10^{-3}}$

$b = \frac{12.5}{0.5}\sqrt{3.996*10^{-3}} = 1.58 kg s^{-1}$

b)$x = x_m e^{\frac{-bt}{2m}}$

   $x = x_m e^{-1.58t}{1} = x_m e^{-1.58t}$    where x_m is initial amplitude

c) critical damping amplitude

$c_c =2\sqrt{km} = 2\sqrt{12.5*.5} = 5 kg s^{-1}$