Newton's Laws question : On springs and pulley system

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SUMMARY

The discussion focuses on the conditions for a particle attached to a spring on a rough inclined plane to oscillate. It establishes that for oscillation to occur, the coefficient of friction, μ, must be less than 1/3 times the tangent of the angle of elevation, θ. The equilibrium position is defined as x_e = (mg(sin θ + μ cos θ)) / k, where k is the spring constant. The analysis concludes that if friction dissipates all the energy before reaching this equilibrium position, oscillation will not occur.

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A Particle is attached by a string to a point in a rough inclined plane of elevation [itex]\theta[/itex]. Co-efficient of friction is [itex]\mu[/itex]. Originally, the string is unstretched and lay along the line of greatest slope.Prove that the condition for particle to oscillate is [itex]\mu[/itex] < 1/3tan[itex]\theta[/itex].
Note : Tension in the string = [itex]\lambda[/itex].[itex]\Delta[/itex]l/l
[itex]\Delta[/itex]l = Change in length
l = Original length
 
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If friction steals away all the energy of the particle before it reaches at most an equilibrium position of [itex]x_e=\frac{mg(sin \theta +\mu cos\theta)}{k}[/itex], where xe is the distance from the starting position of the particle and k is the spring constant, then there is no oscillation. The equilibrium position is gotten by solving:
[tex]mgsin \theta +\mu (mg cos\theta)-kx_e=0[/tex]
Then solve:
[tex]\mu (mg cos\theta) x_e<\frac{1}{2}k (x_e)^2+mg (x_e sin \theta)[/tex]
by substituting in the expression for xe to get the condition for oscillation.
 

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