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.

Question

Accepted Answer

Period of vibration T = 0.8/4 = 0.2 sec
Frequency of vibration ω = 2π/T = [latex]\frac{2\pi }{0.2} [/latex] = 31.4 rad/sec
Slope of the displacement-time envelope curve = [latex] -\frac{2\mu g}{\pi \omega } [/latex]

or,

[latex] ext{Slope} =\frac{50-5}{0.8} =-\frac{45}{0.8} =-\frac{2\mu g}{\pi \omega } [/latex]

or,

[latex]\mu =\frac{\pi \omega }{2g} \left\lgroup\frac{45}{0.8} ight group =\frac{\pi imes 31.4 imes 45}{2 imes 9810 imes 0.8}[/latex] = 0.2828

The displacement [latex]x_{F}[/latex] is given by

[latex]x_{F} =\frac{\mu g}{\omega ^{2}} =\frac{0.2828 imes 9810}{31.4^{2}} [/latex] = 2.81 mm

It will come to rest when

[latex] x(0)-4 x_{F}n

or,

[latex]n=\frac{x(0)-x_{F}}{4x_{F}} =\frac{50-2.81}{4 imes 2.81} =4.198[/latex] 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 [latex]x_{F}[/latex] × 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.

Question 2.8: A SDOF system with a friction damper vibrates with an initia...

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