A 350 kg mass, constrained to move only vertically, is supported by two springs, each having a spring constant of 250 kN/m. A periodic force with a maximum value of 100 N is applied to the mass with a frequency of 2.5 rad/s. Given a damping factor of 0.125, the amplitude of the vibration is

Respuesta :

Answer:

Explanation:

Expression for amplitude of forced damped oscillation is as follows

A = [tex]\frac{F_0}{\sqrt{m(\omega^2-\omega_0^2)^2+b^2\omega^2} }[/tex]

where

ω₀ =[tex]\sqrt{\frac{k}{m} }[/tex]

ω₀ = [tex]\sqrt{\frac{500000}{350} }[/tex]

=37.8

b = .125 ,

ω = 2.5

m = 350  

A = [tex]\frac{100}{\sqrt{350(37.8^2-2.5^2)^2+.125^2\times2.5^2} }[/tex]

A = 3.75 mm . Ans