Question #81301

A mass of 1 kg is attached to a spring having a stiffness of 3920 N/m. The mass slides on a horizontal surface, the co-efficient of friction between mass and surface being 0.1. Determine the frequency of vibration of the system and amplitude after one cycle if the initial amplitude is 0.25 cm.
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Expert's answer

2018-09-28T06:19:06-0400

Question #81301

A mass of 1kg1\,\mathrm{kg} is attached to a spring having a stiffness of 3920N/m3920\,\mathrm{N/m}. The mass slides on a horizontal surface, the co-efficient of friction between mass and surface being 0.1. Determine the frequency of vibration of the system and amplitude after one cycle if the initial amplitude is 0.25cm0.25\,\mathrm{cm}.

Answer:

The frequency of vibration is given by:


f=12πkm,f = \frac{1}{2\pi} \sqrt{\frac{k}{m}},


where kk is spring stiffness and mm is the mass. Thus


f=12π39201=9.965Hz.f = \frac{1}{2\pi} \sqrt{\frac{3920}{1}} = 9.965\,\mathrm{Hz}.


Based on the coefficient of friction μ=0.1\mu = 0.1, we can determine the frictional force as follows


F=μmg=0.119.81=0.981N.F = \mu m g = 0.1 \cdot 1 \cdot 9.81 = 0.981\,\mathrm{N}.


Since, the reduction in amplitude per cycle is given by


ΔA=4Fk=40.9813920=0.0010m=0.1cm,\Delta A = \frac{4F}{k} = \frac{4 \cdot 0.981}{3920} = 0.0010\,\mathrm{m} = 0.1\,\mathrm{cm},


the amplitude after one cycle equal to 0.250.1=0.15cm0.25 - 0.1 = 0.15\,\mathrm{cm}.


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