A mass of 2kg oscillates on a spring with force constant 50 N/m. By what factor does the frequency of oscillation decrease when a damping force with constant b = 12 is introduced? 

We know that,

mass m = 2kg

force constant = 50 N/m

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}$

${\omega }_n=5 rad{s}^{-1}$

Now the damped frequency, 

${\omega }_d=\sqrt{{{\omega }_n}^2-{\left(\frac{b}{2m}\right)}^2}$

${\omega }_d=\sqrt{5^2-{\left(\frac{12}{4}\right)}^2}$

${\omega }_d=4 rad/s$

Percentage reduction = $\frac{5-4}{5}\times 100=20\%$

Hence the correct answer is decreases by 20%.