Friction Damping 1: UNIZOR.COM - Physics4Teens - Waves - Mechanical Oscillation
Channel: Zor Shekhtman on UNIZOR Education
Date: 2021-05-23

UNIZOR.COM - Physics through a Looking Glass of Mathematician
Notes to a video lecture on http://www.unizor.comFriction Damping 1 - Energy ApproachThe oscillations that gradually diminish their amplitude are called damped oscillations.Let's assume that object of mass m oscillates horizontally along the X-axis on a spring with elasticity k, while lying on a horizontal surface in the gravitational field with free fall acceleration g and experiences the constant force of friction as it moves left and right on the spring.In this case there is a constant friction force acting against the movement equaled toFf = μ·m·gwhere μ is the coefficient of friction.For simplicity, we will not differentiate friction at motion (kinetic friction) from friction at rest (static friction) and assume they are both equal to the expression above.Let x(t) be a horizontal displacement of the object from its neutral position on a spring, where a spring is not stretched or squeezed. Initial conditions at time t=0=t0 are:x(0) = a greater than 0x'(0) = 0For oscillation to start, we have to assume that the force of a spring Fs stretched by an initial distance a (according to Hooke's Law, Fs=−k·a) is greater by absolute value than the friction force Ff=μ·m·g.So, the necessary condition for a movement to start isk·a greater than μ·m·gora greater than μ·m·g/k = λwhere λ is a constant that combines the characteristics of a spring's elasticity k, object's mass m and friction μ·g, a compounded characteristic of an oscillating system as a whole.Let's call this characteristic a system's critical distance.Two points, x=λ and x=−λ, that are positioned exactly at critical distance from the neutral point x=0, we will call critical points with the former being a critical point of stretching and the latter - critical point of squeezing.If, holding an object, we stretch a spring by distance a and then let it go, to start a movement, an object should be at a displacement from the neutral position greater by absolute value than critical distance λ. Otherwise, it will remain at rest, the pulling force of a spring would not be sufficient to overcome the friction force.Let's assume that our initial spring stretch a is greater than critical distance λ, that is beyond the critical point of stretching. We expect that the pulling force of a spring will overcome friction, and an object will start moving in the negative direction of the X-axis towards the neutral position and, possibly, beyond it. As it moves, its potential energy is partially spent on friction.Sooner or later it will stop, not reaching the point of x=−a on the opposite side from the neutral point, because, if it did, its potential energy would be the same as in the beginning of motion, which is impossible since energy is partially spent to do some work against the force of friction.To find the stop point b, we will use the energy considerations.When an object started its motion at time t0=0 at position x(t0)=a, its total (potential) energy, as we discussed in the previous lecture, wasU(t0) = k·a²/2When, after moving in negative direction along X-axis, it stops at coordinate b (positive or negative) at some time t1, its total (potential) energy isU(t1) = k·b²/2Since constant friction force Ff=μ·m·g was acting all the time against the object's motion, from the Law of Energy Conservation follows that the work performed by this friction force during the time of motion W[a,b]=Ff·(a−b) equals to a difference in potential energies at the beginning and at the end of motion:U(t0) − U(t1) = W[a,b]From this equation follows:k·a²/2 − k·b²/2 = μ·m·g·(a−b)(a+b)/2 = λthat is, the critical point is a midpoint between beginning and end of motion.Analogously, if we squeeze a spring beyond the critical point of squeezing x=−λ and let it go, the object will start moving to a positive direction along the X-axis and will stop at a point symmetrical to the starting point relatively to the critical point of squeezing.We can resolve the last equation for b to find exact position where our object stops its motionb = 2λ − a = −(a − 2λ)1. If 0 greater than a≤λ, the object will stay at rest.2. If λ greater than a≤2λ, the object stops at distance 0≤b less than λ from the neutral point. 3. If 2λ greater than a≤3λ, the object stops at distance −λ≤b less than 0 from the neutral point. 4. If 3λ greater than a≤4λ, the object stops at distance −2≤b less than −λ from the neutral point. At this point it makes sense to consider this situation as the new beginning, and use the initial position b=2λ−a instead of a, repeating all the calculations, coming with a point x=c, where our object stops the next time, that is symmetrical to x=b relatively to the critical point of squeezing x=−λ:(b+c)/2 = −λc = −2λ − bc = −2λ − (2λ−a) = a − 4λThe last case demonstrates that the absolute value of deviation of an object from the neutral position, the amplitude of its oscillation, is diminishing by 2λ in each direction every time.