Two masses, a rope and a spring

[Nexus99]

Homework Statement

Two blocks of mass M1 and M2 are hung as shown in the figure: $M_1$ is hung from the ceiling through an inextensible rope of negligible mass, while $M_2$ is hung on $M_1$ by an ideal spring of elastic constant k and rest length $L_0$. At the initial moment (t = 0) the masses are stationary and the spring is compressed. For t = 0 the distance between $M_1$ and $M_2$ is equal to $\frac{L_0}{2}$. In the next motion, $M_2$ goes down to reach a maximum distance $L_{max}$ from $M_1$. Calculate
1. The magnitude of the tension T for t=0
2. The acceleration of $M_2$ for t = 0.
3. The value of Lmax and the corresponding magnitude of the tension T
4. The maximum speed reached by $M_2$ during the descent.

M1 = 1.20 kg, M2 = 0.25 kg, k = 30 N/m, L0 = 0.20 m.

Relevant Equations

conservation of energy, harmonic oscillator etc.

I tried in this way:
1) Considering a reference axis oriented downwards:
$M_1:$ $-T + M_1g - k \frac{L_0}{2} = 0$
$T = M_1g - k \frac{L_0}{2}$

2)
$M_2:$ $M_2g + k \frac{L_0}{2} = M_2 a$
$a = g + \frac{k}{M_2} \frac{L_0}{2}$

3) $M_2g \frac{L_0}{2} + \frac{1}{2} k \frac{L_0^2}{4} = M_2gL_{max} + \frac{1}{2} k (L_{max} - L_0)^2$
$L_{max} = \frac{3}{2}L_0 - \frac{2gM_2}{k}$

And
$T = M_1g + k(L_{max} - L_0)$

4) Maximum velocity is reached when the spring is at equilibrium:
$M_2g \frac{L_0}{2} + \frac{1}{2}k (\frac{L_0}{2})^2 = M_2gL_0 + \frac{1}{2}M_2 v_{max}^2$
But solving this equation i got a complex value for $v_{max}$:
$v_{max} = \sqrt{\frac{kL_0^2}{4M_2} - g L_0}$
Where am i wrong?

 

[Nexus99]

I got a new idea for $v_{max}$
$v_{max} = A \omega$ where A is the amplitude of the harmonic motion
$A = L_{equilibrium} - (L_i) = L_0 - \frac{M_2g}{k} - (\frac{L_0}{2} - L_0) = \frac{3}{2}L_0 - \frac{M_2g}{k}$
$v_{max} = (\frac{3}{2}L_0 - \frac{M_2g}{k}) \sqrt{\frac{k}{M_2}}$
Why my first solution was wrong?

 

[haruspex]

Why my first solution was

In your energy equation for 3, for the gravitational PE on the left you seem to have taken the lower mass in the relaxed position of the spring as the zero height datum. Not sure what you have used as datum on the right. At Lmax, what is the height of that mass relative to your zero datum?

 

[Nexus99]

I took the lower mass position for t=0 as zero for gravitation PE. Same for RHS where $M_2$ distance from $M_1$ seems to be $L_{max}$

 

[haruspex]

I took the lower mass position for t=0 as zero for gravitation PE. Same for RHS where $M_2$ distance from $M_1$ seems to be $L_{max}$

But on the RHS you have it positive, which means Lmax above the reference zero.
And if you took the t=0 position of M₂ as the reference zero then the GPE term on the LHS would be zero.

 

[Nexus99]

Sorry i took the higher mass position as zero

 

[haruspex]

Sorry i took the higher mass position as zero

But the signs are still wrong. You have the GPE increasing from $M_2g \frac{L_0}{2}$ to $M_2gL_{max}$:

3) $M_2g \frac{L_0}{2} + \frac{1}{2} k \frac{L_0^2}{4} = M_2gL_{max} + \frac{1}{2} k (L_{max} - L_0)^2$

This is also responsible for the imaginary value for velocity in (4).

 

[Nexus99]

So should be:
$-M_2g \frac{L_0}{2} + \frac{1}{2} k (\frac{L_0}{2})^2 = -M_2gL_{max} + \frac{1}{2} k (L_{max} - L_0)^2$

and

$-M_2g \frac{L_0}{2} + \frac{1}{2} k (\frac{L_0}{2})^2 = -M_2g L_0 + \frac{1}{2} m (v_{max})^2$

Look fine, thanks

 

[haruspex]

So should be:
$-M_2g \frac{L_0}{2} + \frac{1}{2} k (\frac{L_0}{2})^2 = -M_2gL_{max} + \frac{1}{2} k (L_{max} - L_0)^2$

and

$-M_2g \frac{L_0}{2} + \frac{1}{2} k (L_0)^2 = -M_2g L_0 + \frac{1}{2} m (v_{max})^2$

Look fine, thanks

Better, but another problem in the second equation.
As you wrote, "4) Maximum velocity is reached when the spring is at equilibrium:", but that is not at L₀. L₀ is the relaxed length.

 

[Nexus99]

You're right, i have to add $\frac{1}{2} k (L_{eq} - L_0)^ 2$ in the RHS