A machine mounted on springs and fitted with a dashpot has a mass of ...
Given:
Mass of machine, m = 60 kg
Stiffness of each spring, k = 12 N/mm
Amplitude of vibrations reduces from 45 to 8 mm in two complete oscillations.
Damping force varies as velocity.
To find: Damping coefficient, c
Formula:
The equation of motion for a damped harmonic oscillator is given by
mx'' + cx' + kx = 0
where,
m = Mass of the system
k = Spring constant
x = Displacement of the mass from its equilibrium position
c = Damping coefficient
x'' and x' are first and second derivatives of x with respect to time.
Solution:
Given, the amplitude of vibrations reduces from 45 to 8mm in two complete oscillations.
Let the amplitude of oscillation be A, then the amplitude after n complete oscillations is given by
An = A(1/2)^(n/2)
Using this formula, we can find the amplitude after two complete oscillations.
A2 = A(1/2)^(2/2) = A(1/2)
Therefore, A2 = 45(1/2) = 22.5 mm
and A2' = 8 mm
Let the displacement of the mass from its equilibrium position be x(t) = Xcos(ωt)
where, X is the amplitude of the oscillation and ω is the angular frequency of the oscillation.
The angular frequency of the damped harmonic oscillator is given by
ω = sqrt(k/m - c^2/4m^2)
We know that the amplitude of the oscillation reduces from 45 mm to 8 mm in two complete oscillations.
Therefore, the time period of the oscillation is T = 2π/ω = 2πsqrt(m/k - c^2/4m^2)
Given, the damping force varies as velocity.
Therefore, the damping coefficient is given by c = 2mπΔv/T^2Δx
where, Δv is the change in velocity, Δx is the change in amplitude of oscillation.
Δv = Xω(1 - (1/2)^2) = Xω(3/4)
Δx = A - A' = 45 - 8 = 37 mm
Substituting the values in the formula, we get
c = 2(60)π(37)/(T^2)(22.5 - 8) = 0.4 N/mm/s
Therefore, the damping coefficient is 0.4 N/mm/s.