Please show and explain your work, thanks! 6. A damped harmonic oscillator consi

Question

Please show and explain your work, thanks!

6. A damped harmonic oscillator consists of a block of mass m, attached to a spring of stiffness k, sliding on a floor with kinetic friction coefficient . Note that the damping force in this case is not proportional to velocity but instead comes from the friction coefficient. The block is initially extended a distance xo to the right (past equilibrium) and released from rest. (a) Set up the differential equation and solve for the motion during the first half-period while the block is moving to the left. (Be careful of signs; note that the frictional force always points to the right during this part of the motion.) (b) Find the maximum compression of the spring before the block turns around. By analogy, find the maximum extension of the spring at the completion of the first period. Therefore show that the amplitude decreases linearly with time.

Explanation / Answer

6. given

A damped harmonic osscilator consist of a block of mass m, attached to a spring of stiffness k

sliding on tyhe flook with kinetic fricion, fk = u*mg

x(0) = xo

v(0) = 0

hence

the equation of motion is

mx" + kx + u*mg = 0

so the solution can be of the form

x(t) = -c1*m*exp(-kt/m)/k + c2 - u*g*m*t/k

now

x(0) = xo

hence

-c1*m/k + c2 = xo

x'(0) = 0

c1 = u*g*m/k

hence

c2 = xo + u*g*m^2/k^2

hecne

x(t) = -u*g*m^2*exp(-kt/m)/k^2 + xo + ug*m^2/k^2 - ugmt/k

x(t) = xo + ug(m^2/k^2)(1 - exp(-kt/m)) - ugmt/k

b. maximum compression before the blockl tuens around = x

then

from conservation of energy

0.5k*xo^2 = 0.5*kx^2 + u*mg*(x + xo)

x^2 + 2(umg)x/k + (2umg*xo/k - xo^2) = 0

x = (-umg/k + sqrt(u^2m^2g^2/k^2 - 2umg*xo/k + xo^2))
as we can see, x/x1 < 1
hence
the amplitude decreases with time

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