Two objects attached with a spring

[giokrutoi]

not the spring constant, right?

no I have spring constant 20n/m.i forgot to mention it in early posts I said that k is given

 

[ehild]

hey I think I figured out everything
look
we can calculate maximal speed of m1 at the release point from conservation of energy all the potential energy is converted in kinetic of m1
then this v is maximal for m1 thing is that m1's coordinate dependence on time look's so
y=XmaxSINUS OMEGA t were we know omega

That Vmax is valid in the ground frame of reference. After m2 is released, the whole system translates with the velocity of the center of mass, so both objects move with that velocity + the velocity of their vibration.
You turned to the center of mass frame of reference, where the blocks oscillate , but the maximum speed of oscillation is different from Vmax you got previously. You have to subtract the velocity of the center of mass.
If you got it, you can use that the maximum acceleration is maximum speed multiplied by omega. that you know already.

 

[giokrutoi]

That Vmax is valid in the ground frame of reference. After m2 is released, the whole system translates with the velocity of the center of mass, so both objects move with that velocity + the velocity of their vibration.
You turned to the center of mass frame of reference, where the blocks oscillate , but the maximum speed of oscillation is different from Vmax you got previously. You have to subtract the velocity of the center of mass.
If you got it, you can use that the maximum acceleration is maximum speed multiplied by omega. that you know already.

yes that's exactly what I was saying
thanks a lot. For waisting your time on. Helping me

 

[haruspex]

yes that's exactly what I was saying
thanks a lot. For waisting your time on. Helping me

Just to check, the max acceleration I got for m₂ is $\frac{kx_1}{\sqrt{m_2(m_1+m_2)}}$

 

[ehild]

Just to check, the max acceleration I got for m₂ is $\frac{kx_1}{\sqrt{m_2(m_1+m_2)}}$

It depends what x₁ means. It should be the initial compression, 0.1 m.

 

[ehild]

It depends what x₁ means. It should be the initial compression, 0.1 m.

I outline the energy method suggested by jbriggs and me :
ΔL = 0.1 m is the initial compression of the spring.
V1 is the maximum speed of block1 after it was released, but block2 still fixed.
The total energy of the system is 1/2 m₁V₁²=1/2 k (ΔL)²
$V_1 = \sqrt \frac {k}{m_1}\Delta L$
Speed of the CoM $V_{CoM}= \frac {m_1V_1}{m_1+m_2}$
energy of translation $E_{tr}=1/2 (m_1+m_2 ) \left( \frac {m_1V_1}{m_1+m_2} \right)^2$.
Let be S the maximum distortion of the spring after both blocks were released.
The spring energy is $E_s =E-E_{tr}= 1/2 V_1^2 \frac {m_1m_2}{m_1+m_2}= 1/2 k S^2$
$S=V_1\sqrt {\frac {m_1m_2}{k(m_1+m_2)}}$ and a₂=KS/m₂.

 

[haruspex]

I outline the energy method suggested by jbriggs and me :
ΔL = 0.1 m is the initial compression of the spring.
V1 is the maximum speed of block1 after it was released, but block2 still fixed.
The total energy of the system is 1/2 m₁V₁²=1/2 k (ΔL)²
$V_1 = \sqrt \frac {k}{m_1}\Delta L$
Speed of the CoM $V_{CoM}= \frac {m_1V_1}{m_1+m_2}$
energy of translation $E_{tr}=1/2 (m_1+m_2 ) \left( \frac {m_1V_1}{m_1+m_2} \right)^2$.
Let be S the maximum distortion of the spring after both blocks were released.
The spring energy is $E_s =E-E_{tr}= 1/2 V_1^2 \frac {m_1m_2}{m_1+m_2}= 1/2 k S^2$
$S=V_1^2\sqrt {\frac {m_1m_2}{k(m_1+m_2)}}$ and a₂=KS/m₂.

Yes, that looks to be what I did too. No need to get into the SHM details.

 

[giokrutoi]

thanks very much