Answer to Question #135988 in Mechanical Engineering for Ashi

Question

A 800 kg machine is mounted on four identical springs of total spring constant k = 1.5 x 106 N/m and negligible damping. The machine is subjected to a harmonic external force of amplitude Fo = 490 N and frequency of 180 rpm. The amplitude of the motion of the machine

Answer 1

$A_0=\frac{F_0}{m\sqrt{(\omega_0^2-\omega^2)^2+4\beta^2\omega^2}}$

$180(rpm)=3(c^{-1})$

$\omega=2\pi\nu=2\cdot3.14\cdot3=18.84(rad/s)$

$\omega_0^2=k/m=1.5\cdot10^6/800=1875(rad^2/s^2)$

For negligible damping $A_0\approx\frac{F_0}{m\sqrt{(\omega_0^2-\omega^2)^2}}=\frac{F_0}{m(\omega_0^2-\omega^2)}\to$

$A_0=\frac{F_0}{m(\omega_0^2-\omega^2)}=\frac{490}{800\cdot(1875-18.84^2)}=4\cdot10^{-4}(m)$