By what factor time constant ‘T’ should be multiplied so that damping ratio is reduced from 0.6 to 0.4.

The closed loop transfer function is,

T.F=$\frac{(C(S))}{(R(S))}$

=$\frac{(G(S))}{(1+G(S)H(S))}$, H(S)=1

=$\frac{(\frac{k}{(S(TS+1))})}{(\frac{1+k}{(S(TS+1))} x1)}$

=$\frac{k}{(TS^2+S+k)}$

=$\frac{(\frac{k}{T})}{(\frac{S^2+1}{T S}+\frac{k}{T})}$

Comparing with the standard form of the equation

$\frac{(w_n^2)}{(S^2+2ξw_n S+w_n^2 )}$

$w_n^2$ =$\frac{k}{T}$

Therefore, $w_n$=${\sqrt{\frac{k}{T}}}$

$2ξw_n$=$\frac{1}{T}$

Therefore, ξ=$\frac{1}{2}{\sqrt{kT}}$

Let $ξ_1$=0.3 and $ξ_2$=0.8

$ξ_1$=$\frac{1}{2}{\sqrt{kT}}$

0.3=$\frac{1}{2}{\sqrt{kT}}$……….(1)

Let for $ξ_2$, k=$k_1$

Therefore, 0.8=$\frac{1}{2}{\sqrt{(k_1 T)}}$…..(2)

From equation (1) & (2)

${\sqrt{kT}}$=$\frac{1}{(2 x 0.3)}$ =1/0.6

${\sqrt{k_1 T}}$=$\frac{1}{(2 x 0.8)}$

=1/1.6

$\frac{\sqrt{kT}}{\sqrt{(k_1 T)}}$=$\frac{1}{0.6}$ x $\frac{1.6}{1}$

=1.6/0.6

$\frac{k}{k_1}$ =$\frac{2.56}{0.36}$

=7.11

$k_1$=0.1406k

Hence, the multiplying factor is 0.1406k.