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A vibrometer having a natural frequency of 4 rad/s and damping ratio of 0.2 is attached to a structure that performs a harmonic motion
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If the difference between the maximum and the minimum recorded values in 8mm, Find the amplitude of motion of the vibrating structure when its frequency is 40 rad/sec.

$w_n = 4r/sec$

$\xi$ = 0.2

W = 40 rad/sec

2Z = 0.0008 m

= 0.004m

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Y = ?

$\frac{Z}{Y} = \frac{ (40/4)^2}{\sqrt{ (1- (\frac{40}{4})^2)^2 + (2 \times 0.2 \times \frac{40}{4})^2}}$

$Y = 3.96 \times 10^{-3} m$

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