The pointer of the vibration measuring instrument is observed to move between the 0.1 and0.3 marks on the vertical scale, when subject to a displacement alternating at a frequency of100 rad/s. What would the excursion be if the forcing frequency were doubled

Step-by-step explanation:

Given points moves b/w 0.1 of 0.3 marks Hence amplitude is 0.2

Also , frequency (w) = 100 rad/s.

Natural frequency (wn) = √K/m

K = 20 N/mm x 1000mm/1m = 20000N/M

W = 1000mm/1m = 20,000N/M

FormulaX = FolK/1 - (w/wn)²

0.2 = b/20000/1 - (100/50)²

Given forcing frequency was doubled, W1 = 2 x 100=200rad/s

X1 =b/20,000/1 - (200/20)²

0.2/11 = 1 - (200/50)²/1 - (100/50)² = 0.2/11 = -15^5– 3

X1 = 0.04

the answer is b.

step-by-step explanation: