Driven, damped, simple harmonic oscillator spring driving friction Consider a bl

Question

Driven, damped, simple harmonic oscillator spring driving friction Consider a block of mass m that slides along a surface (in 1 dimension) attached to a wall by a spring The spring obeys Hooke's law, exerting a force on the block: F resting position of the block when no other forces act. Friction between the surface and the block results in a damping force. Fdamp--DX. The block is also acted on by a driving force that oscillates in time: Fdrive - Fo cos(wt). You may not simply refer to book equations for your solutions. You must work them out. Fspringkx, where x-0 is defined a CASE 1: weakly damped, non-driven. Assume Fo = 0 and b

Explanation / Answer

given mass = m

spring constant = k

Fdamp = -bx'

Fdrive = Focos(wt)

CASE 1 : Fo = 0, b < 2*sqroot(km)

a. equation of motion of the block

mx" + bx' + kx = 0

b. let x = Ae^(-lambda*t)

m*lambda^2 - b*lambda + k = 0

solving for lambda

lambda = (b +- sqroot(b^2 - 4mk))/2m

now, b^2 < 4mk hence

this is weakly damped motion

so the solution can be written as

x = e^(-gamma*t)cos(w1t - phi)

gamma = b/2m

w1 = sqroot(wo^2 - gamma^2)

wo = sqroot(k/m)

hence

x = Ae^(-bt/2m)cos(t*sqroot(k/m - b^2/4m) - phi)

x' = xo at t = 0

x = 0 at t = 0

cos(phi) = 0

phi = 90 deg

hence solution becomes

x = Ae^(-bt/2m)sin(t*sqroot(k/m - b^2/4m))

x' = A*e^(-bt/2m)[-b*sin(t*sqroot(k/m - b^2/4m))/2m + sqroot(k/m - b^2/4m)cos(t*sqroot(k/m - b^2/4m)) ]

x' = vo = A*[sqroot(k/m - b^2/4m) ]

A = vo/sqroot(k/m - b^2/4m)

hence

x = vo*e^(-bt/2m)sin(t*sqroot(k/m - b^2/4m))/sqroot(k/m - b^2/4m)

c. the motion is underdamped with the amplitude decreasing in successive oscsilations

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