Conservation of momentum and SHM

[erisedk]

Homework Statement

Two point masses m1 and m2 are coupled by a spring of spring constant k and uncompressed length L₀. The spring is fully compressed and a thread ties the masses together with negligible separation between them. The tied assembly is moving in the +x direction with uniform speed v₀. At a time, say t = 0, it is passing the origin and at that instant the thread breaks. The masses, attached to the spring, start oscillating. The displacement of mass m1 is given by x₁ (t) = v₀t - A(1-cos(ωt)) where A is a constant. Find (i) the displacement x₂(t) of mass m2, and (ii) the relationship between A and L₀.

Homework Equations

The Attempt at a Solution

First part is easy. Using
$x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}$
and substituting $x_{cm} = v_0t$ and $x_1= v_0t - A(1-\cos{ωt})$
we get $x_2 = v_0t + \dfrac{m_1}{m_2}.A(1-\cos{wt})$

However, I'm not sure what to do for part (ii). I suppose it involves using the energy equation, but that isn't really working out because of the $t$ (time). I think we might have to minimise or maximise something, in any case, I'm not sure how to proceed. Please help.

 

[haruspex]

Homework Statement

Two point masses m1 and m2 are coupled by a spring of spring constant k and uncompressed length L₀. The spring is fully compressed and a thread ties the masses together with negligible separation between them. The tied assembly is moving in the +x direction with uniform speed v₀. At a time, say t = 0, it is passing the origin and at that instant the thread breaks. The masses, attached to the spring, start oscillating. The displacement of mass m1 is given by x₁ (t) = v₀t - A(1-cos(ωt)) where A is a constant. Find (i) the displacement x₂(t) of mass m2, and (ii) the relationship between A and L₀.

Homework Equations

The Attempt at a Solution

First part is easy. Using
$x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}$
and substituting $x_{cm} = v_0t$ and $x_1= v_0t - A(1-\cos{ωt})$
we get $x_2 = v_0t + \dfrac{m_1}{m_2}.A(1-\cos{wt})$

However, I'm not sure what to do for part (ii). I suppose it involves using the energy equation, but that isn't really working out because of the $t$ (time). I think we might have to minimise or maximise something, in any case, I'm not sure how to proceed. Please help.

Energy should do it. Please post your working.

 

[erisedk]

Sorry. I forgot to post the working.
Conserving energy at t=0 and t=t,

$\dfrac{1}{2}.k{L_0}^2 + \dfrac{1}{2}(m_1 + m_2){v_0}^2 = \dfrac{1}{2}k{(L_0 - |x_2 - x_1|)}^2 + \dfrac{1}{2}m_1{v_1}^2 + \dfrac{1}{2}m_2{v_2}^2$

where $v_1 = v_0 - Aω\sin{ωt}$ and $v_2 = v_0 + A\dfrac{m_1}{m_2}ω\sin{ωt}$

This is just terrible.
So, I'm thinking there is definitely some optimisation involved, something like $\sin{ωt} = 0$ or $\cos{ωt} = 0$ but I'm not seeing what.

 

[haruspex]

Sorry. I forgot to post the working.
Conserving energy at t=0 and t=t,

$\dfrac{1}{2}.k{L_0}^2 + \dfrac{1}{2}(m_1 + m_2){v_0}^2 = \dfrac{1}{2}k{(L_0 - |x_2 - x_1|)}^2 + \dfrac{1}{2}m_1{v_1}^2 + \dfrac{1}{2}m_2{v_2}^2$

where $v_1 = v_0 - Aω\sin{ωt}$ and $v_2 = v_0 + A\dfrac{m_1}{m_2}ω\sin{ωt}$

This is just terrible.
So, I'm thinking there is definitely some optimisation involved, something like $\sin{ωt} = 0$ or $\cos{ωt} = 0$ but I'm not seeing what.

It is not a question of optimisation. You need to pick a particular time t to compare with t=0.

 

[Shivam aditya]

(ii) maximum compression = maximum stretching = L
So max. gap btw. x1 and x2 = max. Stretch + natural length = L + L = 2L
x2 - x1 = Am1/m2(1-coswt) + A(1-coswt)
(x2 - x1) max. =» 2A(m1/m2) + 2A = 2L
L = A(m1 + m2)/m2

 

[PeroK]

(ii) maximum compression = maximum stretching = L
So max. gap btw. x1 and x2 = max. Stretch + natural length = L + L = 2L
x2 - x1 = Am1/m2(1-coswt) + A(1-coswt)
(x2 - x1) max. =» 2A(m1/m2) + 2A = 2L
L = A(m1 + m2)/m2

The original post was from 3 years ago!