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.
Expert's answer
Question #81301
A mass of 1kg is attached to a spring having a stiffness of 3920N/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 0.25cm.
Answer:
The frequency of vibration is given by:
f=2π1mk,
where k is spring stiffness and m is the mass. Thus
f=2π113920=9.965Hz.
Based on the coefficient of friction μ=0.1, we can determine the frictional force as follows
F=μmg=0.1⋅1⋅9.81=0.981N.
Since, the reduction in amplitude per cycle is given by
ΔA=k4F=39204⋅0.981=0.0010m=0.1cm,
the amplitude after one cycle equal to 0.25−0.1=0.15cm.
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