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Question

A mass of 50 kg is suspended from a spring of stiffness 10 kN/m. It is set into oscillations and it is observed that two successive oscillations have amplitudes 10 mm and 1 mm. Determine the damping ratio.

A
0.232
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B
0.344
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C
0.392
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D
0.446
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Solution

The correct option is B 0.344
Let us consider two oscillations , one occuring m cycles after the first oscillation.
Amplitude of first oscillation at time t is given by x1=Aeδω0t .....(1)
where δ=γω0
Amplitude of second oscillation after m cycles (time mT) is given by x2=Aeδω0(t+mT) ......(2)
From (1) and (2), we can write that,
x1x2=eδω0mT
δω0mT=ln(x1x2) ......(3)
Here, δ is known as damping ratio.
For successive amplitudes, m=1
δω0T=ln(101)=2.3
δω0(2πω)=2.3 ....(4)
But we know that, ω=ω20γ2
From this we get, ω=ω01δ2
where δ=γω0
From (4) we can write that,
2πδ1δ2=2.3
Squaring on both sides ,
39.4 δ2(1δ2)=5.29
δ2=0.118δ=0.344
Thus, option (b) is the correct answer.

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