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A vibrating mass of 300 kg mounted on a massless support by a spring of stiffness 40,000 N>m and a damper of unknown damping coefficient is observed to vibrate with a 10-mm amplitude while the support vibration has a maximum amplitude of only 2.5 mm (at resonance). Calculate the damping constant and the amplitude of the force on the base.

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Answer:

400 N

Step-by-step explanation:


\text { Given: } m=300 \mathrm{~kg}, k=40,000 \mathrm{~N} / \mathrm{m}, \omega_(b)=\omega_(n)(r=1), X=10 \mathrm{~mm}, Y=2.5 \mathrm{~mm} .

Find damping constant


(X)/(Y)=\left[(1+(2 \zeta r)^(2))/(\left[\left(1-r^(2)\right)^(2)+(2 \zeta r)^(2)\right.)\right]^(1 / 2) \\ \left.(10)/(2.5)=(\left\lceil 1+4\zeta^(2)\right]^(1 / 2))/(4 \zeta^(2))\right] \\ 16=(1+4 \zeta^(-2))/(4 \zeta^(2)) \\ \zeta^(2)=(1)/(60)=(c^(2))/(4 k m) \\ c=\sqrt{(4(40,000)(300))/(60)} \\ c=894.4 \mathrm{~kg} / \mathrm{s}

Amplitude of force on base:


F_(T)=k Y r^(2)\left[(1+(2 \zeta r)^(2))/(\left(1-r^(2)\right)^(2)+(2 \zeta r)^(2))\right]^(1 / 2)

substituting the values in above formula we get

F_T = 400 N

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