Question 2.8: A SDOF system with a friction damper vibrates with an initia...
A SDOF system with a friction damper vibrates with an initial displacement of 50 mm. It completes 4 cycles in 0.8 second when its amplitude reduces to 5 mm. Estimate the coefficient of friction and when does the system come to rest.
Period of vibration T = 0.8/4 = 0.2 sec
Frequency of vibration ω = 2π/T = \frac{2\pi }{0.2} = 31.4 rad/sec
Slope of the displacement-time envelope curve = -\frac{2\mu g}{\pi \omega }
or,
\text{Slope} =\frac{50-5}{0.8} =-\frac{45}{0.8} =-\frac{2\mu g}{\pi \omega }
or,
\mu =\frac{\pi \omega }{2g} \left\lgroup\frac{45}{0.8} \right\rgroup =\frac{\pi \times 31.4\times 45}{2\times 9810\times 0.8} = 0.2828
The displacement x_{F} is given by
x_{F} =\frac{\mu g}{\omega ^{2}} =\frac{0.2828\times 9810}{31.4^{2}} = 2.81 mm
It will come to rest when
x(0)-4 x_{F}n<x_{F} and velocity is zero.
or,
n=\frac{x(0)-x_{F}}{4x_{F}} =\frac{50-2.81}{4\times 2.81} =4.198 cycles
It means the system will come to rest at the beginning of next half cycle, that is, 4.5 cycles or at 0.9 sec. The displacement at 0.9 sec will be equal to
x(0) – 4 x_{F} × 4.5 = 50 – 4 × 2.81 × 4.5 = -0.58 mm
The minus sign indicates that 0.58 mm is in a direction opposite to that of the initial displacement.