ma = ? kx ? bv The figure shows a critically damped oscillator. Initially, at ti

Question

ma = ?kx ? bv
The figure shows a critically damped oscillator. Initially, at time t = 0, the mass is reseased from rest at position x = A.
(A) Express b in terms of the other parameters (k, m) for a critically damped oscillator.
(B) Derive the position as a function of time, x(t). (Explain your work.)
(C) Determine the maximum magnitude of the force, for t ? 0.

ma = ?kx ? bv The figure shows a critically damped oscillator. Initially, at time t = 0, the mass is reseased from rest at position x = A. Express b in terms of the other parameters (k, m) for a critically damped oscillator. Derive the position as a function of time, x(t). (Explain your work.) Determine the maximum magnitude of the force, for t ? 0.

Explanation / Answer

a)

at critically damped oscillator:

b = sqrt(4 m k)

b)

ma = ?kx ? bv

==> m a + b v + k x = 0

==> m x'' + b x' + k x = 0

suppose: x = e^(s t)

==> x' = s e^(s t)

==> x'' = s^2 e^(s t)

m x'' + b x' + k x = 0

==> m s^2 e^(s t) + b s e^(s t) + k e^(s t) = 0

==> m s^2 + b s + k = 0

wich has roots:

(-b + sqrt(b^2 - 4mk))/(2m) and   (-b - sqrt(b^2 - 4mk))/(2m)  

we have: b = sqrt(4 m k), therefore the roots are:

(-b)/(2m) and   (-b)/(2m)

We have only one exponential solution, so we need to multiply it by t to get the second solution. The basic solutions are "e^(-bt/2m)" and "t e^(-bt/2m)". The general solution is

x(t) = e^(-bt/2m) (c1 + t c2)

therefore:

v(t) = dx/dt = (-b/2m) e^(-bt/2m) (c1 + t c2) + e^(-bt/2m) (c2)

we have x(0) = A:

x(0) = A = e^(-b*0/2m) (c1 + 0 * c2)

==> c1 = A

we have v(0)=0 :

v(0) = (-b/2m) e^(-b*0/2m) (c1 + 0 * c2) + e^(-b*0/2m) (c2) = 0

==> (-b/2m) (c1) + c2 = 0

==> (-b/2m) (A) + c2 = 0

==> c2 = bA/2m

we can now write x(t):

x(t) = e^(-bt/2m) (A + (t b A)/(2m))

c)

v(t) = dx/dt = (-b/2m) e^(-bt/2m) (c1 + t c2) + e^(-bt/2m) (c2)

a(t) = dv/dt = (-b/2m)^2 e^(-bt/2m) (c1 + t c2) + (-b/2m) e^(-bt/2m) (c2) + (-b/2m) e^(-bt/2m) (c2)

==> a(t) = (-b/2m)^2 e^(-bt/2m) (c1 + t c2) + 2 (-b/2m) e^(-bt/2m) (c2)

==> a(t) = (-b/2m)^2 e^(-bt/2m) (A + t (bA/2m)) + 2 (-b/2m) e^(-bt/2m) (bA/2m)

==> da/dt = 0

==> (-b/2m)^3 e^(-bt/2m) (A + t (bA/2m)) + (-b/2m)^2 e^(-bt/2m) (bA/2m) + 2 (-b/2m)^2 e^(-bt/2m) (bA/2m) = 0

==> (A + t (bA/2m)) - A - 2 A = 0

==> t (b/2m) = 2

==> t = 2m/b

==> a(t=2m/b) = (-b/2m)^2 e^(-1) (A + A) + 2 (-b/2m) e^(-1) (bA/2m)

==> a(t=2m/b) = 4 A (b/2m)^2 e^(-1)

F = m a

==> F = 4 m A (b/2m)^2 e^(-1)

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