Question

Solve it please. "?"

Answer 1

`{(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.

Answer 2

`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)`