Answer:
I can't understand your language
Answer:
Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.
B because thermal has to do with temperature and it’s the amount of kinetic and potential energy in and object
I think that the answer would be B or C
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
<u>a) Calculate Natural frequency </u>
Wn = √k/m = 
= 1549.19 rad/s ≈ 246.56 Hz
<u>b) Bandwidth of instrument </u>
W / Wn = 
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