Question

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

Answer 1

Answer:

Explanation:

Expression for amplitude of forced damped oscillation is as follows

A = $\frac{F_0}{\sqrt{m(\omega^2-\omega_0^2)^2+b^2\omega^2} }$

where

ω₀ =$\sqrt{\frac{k}{m} }$

ω₀ = $\sqrt{\frac{500000}{350} }$

=37.8

b = .125 ,

ω = 2.5

m = 350  

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

A = 3.75 mm . Ans