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Note: Please wait a bit for the math formulas to appear.(I use MathJax.js) 1- Problem 1: A spring-mass system on horizontal surface performs damped oscillation due to Coulomb frictional force \[F_{cl}=\mu mg\]. When t = 0 the mass was pulled so that the spring stretches a segment of $x_0=6$ (cm) then it was loosed. It's known that: \[\omega = \sqrt{\frac{k}{m}}= \frac{\pi}{4} = 0.7854 \quad\text{(rad/s)}\] and the length of the way for one direction movement (from left to right or from right to left) decreasing steadily to 0.8 (cm). a- Establish the equation of motion x = x(t) b- Using LaTeX to draw this oscillating graph. 2- Solution: A. Two half periods at the beginning We will use the horizontal x axis in the direction of elongation of the spring. x = 0 is the position of the mass when it is not pulled or compressed. The condition for the mass to stop is: \[\text{velocity }v=0\quad\text{and}\quad\ |x| \le d\quad\text{where }d = \f

Note: Please wait a bit for the math formulas to appear.(I use MathJax.js) Let consider a spring-mass system with the mass sliding on a horizontal surface. The force of dry friction between the mass and the surface is called Coulomb friction. \[F_{cl} = \mu.m.g\] $\mu$ is the coefficient of friction. The half period (0.5*T) is the time for the mass to move in one direction from rightmost position to leftmost position or vice versa from extreme left to extreme right. The period T is the time for the mass to move from the rightmost position to the left and then moving back to the new rightmost position. We can prove that: \[T = \frac{2\pi}{\omega}\quad\text{and}\quad\omega=\sqrt{\frac{k}{m}}\] The minimum stretch d of spring for oscillation to start ($x_0 > d$): \[k.d = F_{cl}=\mu.m.g\quad\to\quad d = \frac{F_{cl}}{k}\] We can prove that: The loss of amplitude in half period is \[2.d = 2.\frac{\mu.m.g}{k}\] The following is the JPG picture of my w