The question is similar to last week’s, except that we will consider how friction may damp the oscillation with time. A block with mass m shown in the drawing is acted on by a spring with spring constant k. The block is pulled distance [x[/0] from equilibrium position (x=0) and released. The block would oscillate back and forth around equilibrium position (x=0). From the last week’s work, we know that if the friction between
the block and the surface that the block is sitting on can be ignored, the block will oscillate forever with block’s position given as x(t)=X0cos(ωt) where ω=√(k/m) is the angular frequency of the oscillation.
Now consider that in reality there is actually friction between the block and the surface, and the friction coefficient is μ<<1. The block is again pulled distance X0 from equilibrium and released. It oscillates many times and eventually comes to rest. Show that the decrease of amplitude is the same for each cycle of oscillations. Find the number of cycles n the mass oscillates before coming to rest.
Hint:You can still write down x(t) = Acos(ωt) but the amplitude (A) decreases with time. Consider the work done by the friction and what it means to the elastic potential energy of the spring.
x(t) = Acos(ωt)
The Attempt at a Solution
I've tried multiple ways today with not much in terms of progress. Below is an example
ΔPe= 1/2(KX0^2) - fAn f = force by friction, n = number of oscillations, A = amplitude
-1/2mv^2 = 1/2(KX0^2) - fAn I then set it equal to the kinetic energy times -1. I did this because the kinetic at it's maximum is equal to the potential energy at its maximum. My reasoning was that the friction would do work on the block. And then I would be able to find the amplitude. I could then find a ratio between the new amplitude and the original one.
That's as far as I got, I tried cancelling some things out and plugging in some more stuff but nothing has worked so far.