Solve it please. "?"

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2 Answers
Apr 13, 2018

#{(m ddot x_1 + k(x_1-x_2) = -F),(2m ddot x_2 + k(x_2-x_1) = 2 F):}#

Applying the Laplace transform

#{((s^2+k/m)X_1-k/mX_2=-1/mF+dot x_1(0)+s x_1(0)), (-k/(2m)X_1+(s^2+k/(2m))X_2 = 1/(2m)F+dot x_2(0)+s x_2(0)):}#

Now assuming

#dot x_1(0) = -u#
#dot x_2(0) = 2 u#

#x_1(0)=x_2(0) = 0#

#((s^2+k/m,-k/m),(-k/(2m),s^2+k/(2m)))((X_1),(X_2)) = ((-1/mF-u),(1/(2m)F + 2u))#

and after the anti transformation gives

#{(x_1 = u t - sqrt(2/(3km))(F+2m u)sin(sqrt((3k)/(2m))t)),(x_2 = u t + sqrt(1/(6km))(F+2m u)sin(sqrt((3k)/(2m))t)):}#

hence

#x_2-x_1 = (sqrt(1/(6km))+sqrt(2/(3km)))(F+2m u)sin(sqrt((3k)/(2m))t)#

The conclusions are left to the reader.

Apr 13, 2018

Answer:

#xi_(max) =sqrt((16 F^2)/(9 k^2) + (6 m u^2)/(k) ) + (4 F)/(3 k)#

Explanation:

For spring extension is #xi = x_B - x_A#, Newton's 2nd Law applied to blocks is:

Block A: #qquad k xi - F = m ddot x_A#

Block B: #qquad 2F - k xi = 2m ddot x_B#

Newton's Law for the system (ignoring spring as it is in the system):

#2F - F = (m_A + m_B) ddot x_(cm) implies F = 3m ddot x_(cm) qquad triangle#

Centre of mass of system:

#x_(cm) = (m_A x_A + m_Bx_B)/(m_A + m_B) = 1/3x_A + 2/3 x_B#, so:

....so:

#((xi),(x_(cm))) = ((-1,1),(1/3,2/3)) ((x_A),(x_B))#

Inverting this:

# ((x_A),(x_B)) = ((-2/3,1),(1/3,1)) ((xi),(x_(cm)))#

and

# ((ddot x_A),(ddot x_B)) = ((-2/3,1),(1/3,1)) ((ddot xi),(ddot x_(cm)))#

Our equations for A& B become:

Block A: #qquad k xi - F = m ddot x_(cm) - 2/3 m ddot xi#

Block B: #qquad 2F - k xi = 2m ddot x_(cm) + 2/3 m ddot xi#

We can eliminate #x_(cm)# using #triangle#:

Block A: #qquad k xi - F = F/3 - 2/3 m ddot xi#

# implies ddot xi + (3k)/(2m) xi = 2 F/m #

This solves as:

#xi(t) = alpha sin(omega t) + beta cos(omega t) + (4 F)/(3 k)# with #omega = sqrt( (3k)/(2m))#

#xi(0) = 0 implies beta = -(4 F)/(3 k) #

#xi' = alpha omega cos(omega t) - beta omega sin(omega t ) #

#xi'(0) = 3u implies alpha omega = 3u #

#implies xi(t) = (3u)/omega sin(omega t) -(4 F)/(3 k) cos(omega t) + (4 F)/(3 k)#

#=Gamma( ((3u)/omega)/Gamma sin(omega t) -((4 F)/(3 k) )/Gamma cos(omega t)) + (4 F)/(3 k)#

where #Gamma = sqrt(((3u)/omega)^2 + ((4 F)/(3 k) )^2) = sqrt(16 F^2 ω^2 + 81 k^2 u^2)/(3k omega)#

#=sqrt(16 F^2 ω^2 + 81 k^2 u^2)/(3k omega) sin(omega t - phi) + (4 F)/(3 k)#

EDIT: Now #omega^2 = (3k)/(2m)# so this simplifies further to:

#=sqrt((16 F^2)/(9 k^2) + (6 m u^2)/(k) ) sin(omega t - phi) + (4 F)/(3 k)#

Max Extension is therefore:

#xi_(max) =sqrt((16 F^2)/(9 k^2) + (6 m u^2)/(k) ) + (4 F)/(3 k)#