Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
a) Calculate Natural frequency
Wn = √k/m = [tex]\sqrt{12000 / 5*10^{-3} }[/tex]
= 1549.19 rad/s ≈ 246.56 Hz
b) Bandwidth of instrument
W / Wn = [tex]\sqrt{1-2(0.4)^2}[/tex]
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs
