Question 10.3.1: (a) Determine the units of the logarithmic decrement (b) If ...

(a) It is clear from Eqn 10.2-2 that the unit of the quantity c(dx/dt) is N, hence

M\frac{d^2x}{dt^2}+c\frac{dx}{dt}+kx=P          (10.2-2)

(units of c) \times (ms^{-1}) = N

=kg \ m \ s^{-2}

then

units of c =(kg \ m \ s^{-2})m^{-1} \ s

=kg \ s^{-1}        (i.e. kilograms per second)

From Eqn 10.3-8,

\begin{matrix} m_1 \\ m_2 \end{matrix}=-\frac{c}{2M}\pm i\frac{1}{2M}\left(4kM-c^2\right)  where  i =√-1

=-\frac{c}{2M}\pm i\omega_d            (10.3-8)

units of \omega_d=\frac{√\left[\left(N \ m^{-1}\right)kg-\left(kg \ s^{-1}\right)^2 \right] }{kg}

=√\left(N \ m^{-1} \ kg^{-1}-s^{-2}\right)

=√\left[(kg \ m \ s^{-2})m^{-1} \ kg^{-1}-s^{-2}\right]=√s^{-2}

=s^{-1}             (radians per second)

then

units of period T_d=1/s^{-1}=s       (seconds)

From Eqn 10.3-16

logarithmic decrement \delta =\frac{cT_d}{2M}         (10.3-16)

units of \delta=\frac{(units \ of \ c)(units \ of \ T_d)}{(units \ of \ M)}

=\frac{(kg \ s^{-1})(s)}{kg}

=\underline{dimensionless}

i.e. as expected, the logarithmic decrement \delta is a number, without units.
(b) From Eqn 10.3-16,

\delta =\frac{cT_d}{2M}    where    T_d=\frac{2\pi }{w_4}    (see Eqn 10-3-14)

and

\omega_d=\frac{(4kM-c^2)^{1/2}}{2M}      (see Eqn 10.3-8)

then

\delta =\frac{2\pi c}{(4kM-c^2)^{1/2}}  and  c = rc_r   (given)

=r\times2√(kM)    (from Eqn 10.3-6)

c_c=2√(kM)    (10.3-6)

Hence

\delta =\frac{2\pi r\left[2√(kM)\right] }{\left(4kM-r^24kM\right)^{1/2} }=\underline{2\pi r(1-r^2)^{-1/2}}

For practical structures, r often lies between a few per cent to 20%, so that \delta \approx 2\pi r.