Respuesta :
The period of the small oscillations that result when the rod is rotated slightly and released is 0.0653s.
Harmonic motion, where the damping force is proportional to the velocity, which is a realistic damping force for a body moving through a fluid is called damped harmonic motion.
If τ=−Cθ, where τ is the torque, θ is the angle of rotation, and C is a constant of proportionality, then the angular frequency of oscillation is ω= sqrtC/I and the period is T=2π/ω=2π sqrtI/C.
where, I is the rotational inertia of the rod. The plan is to find the torque as a function of θ and identify the constant C in terms of identies given in question.
Now using T=2π sqrtI/C, we get
T= 2π sqrtm/3K
putting given quantities,
T= 0.0653s.
The question is incomplete, the complete question is
In the overhead view of above figure, a long uniform rod of mass 0.600kg is free to rotate in a horizontal plane about a vertical axis through its center. A spring with force constant k=1850N/m is connected horizontally between one end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. What is the period of the small oscillations that result when the rod is rotated slightly and released?
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