Question

Two masses m1andm2 re connected by a spring of spring constnt k and are placed on as frictionless horizontal surface. Initially the spring is tretched through a distance x0 when the system is released from rest. Find the distance moved by the tow masses before they again come to rest.

Answer 1

The two masses again come to rest when the spring gets compressed to the same distance $\sf{x_0}$ in total. Let the distance moved by $\sf{m_1}$ and $\sf{m_2}$ during this time be $\sf{x_1}$ and $\sf{x_2}$ respectively, which are opposite to each other.

Thus when the spring is stretched, the masses $\sf{m_1}$ and $\sf{m_2}$ are at distances $\sf{\dfrac{x_1}{2}}$ and $\sf{\dfrac{x_2}{2}}$ respectively, from their initial positions.

Since the net extension is $\sf{x_0,}$

$\displaystyle\longrightarrow\sf{\dfrac{x_1}{2}+\dfrac{x_2}{2}=x_0}$

$\displaystyle\longrightarrow\sf{x_1+x_2=2x_0\quad\quad\dots(1)}$

Here the center of mass of the system has no displacement even the masses have individual displacements.

$\displaystyle\longrightarrow\sf{\bar x=0}$

Displacements of $\sf{m_1}$ and $\sf{m_2}$ are opposite to each other.

$\displaystyle\longrightarrow\sf{\dfrac{m_1x_1+m_2(-x_2)}{m_1+m_2}=0}$

$\displaystyle\longrightarrow\sf{m_1x_1-m_2x_2=0}$

From (1),

$\displaystyle\longrightarrow\sf{m_1x_1-m_2(2x_0-x_1)=0}$

$\displaystyle\longrightarrow\sf{m_1x_1-2m_2x_0+m_2x_1=0}$

$\displaystyle\longrightarrow\sf{\underline{\underline{x_1=\dfrac{2m_2\,x_0}{m_1+m_2}}}}$

Similarly,

$\displaystyle\longrightarrow\sf{\underline{\underline{x_2=\dfrac{2m_1\,x_0}{m_1+m_2}}}}$

Answer 2

Answer:

The two masses again come to rest when the spring gets compressed to the same distance \sf{x_0}x

0

in total. Let the distance moved by \sf{m_1}m

1

and \sf{m_2}m

2

during this time be \sf{x_1}x

1

and \sf{x_2}x

2

respectively, which are opposite to each other.

Thus when the spring is stretched, the masses \sf{m_1}m

1

and \sf{m_2}m

2

are at distances \sf{\dfrac{x_1}{2}}

2

x

1

and \sf{\dfrac{x_2}{2}}

2

x

2

respectively, from their initial positions.

Since the net extension is \sf{x_0,}x

0

,

\displaystyle\longrightarrow\sf{\dfrac{x_1}{2}+\dfrac{x_2}{2}=x_0}⟶

2

x

1

+

2

x

2

=x

0

\displaystyle\longrightarrow\sf{x_1+x_2=2x_0\quad\quad\dots(1)}⟶x

1

+x

2

=2x

0

…(1)

Here the center of mass of the system has no displacement even the masses have individual displacements.

\displaystyle\longrightarrow\sf{\bar x=0}⟶

x

ˉ

=0

Displacements of \sf{m_1}m

1

and \sf{m_2}m

2

are opposite to each other.

\displaystyle\longrightarrow\sf{\dfrac{m_1x_1+m_2(-x_2)}{m_1+m_2}=0}⟶

m

1

+m

2

m

1

x

1

+m

2

(−x

2

)

=0

\displaystyle\longrightarrow\sf{m_1x_1-m_2x_2=0}⟶m

1

x

1

−m

2

x

2

=0

From (1),

\displaystyle\longrightarrow\sf{m_1x_1-m_2(2x_0-x_1)=0}⟶m

1

x

1

−m

2

(2x

0

−x

1

)=0

\displaystyle\longrightarrow\sf{m_1x_1-2m_2x_0+m_2x_1=0}⟶m

1

x

1

−2m

2

x

0

+m

2

x

1

=0

\displaystyle\longrightarrow\sf{\underline{\underline{x_1=\dfrac{2m_2\,x_0}{m_1+m_2}}}}⟶

x

1

=

m

1

+m

2

2m

2

x

0

Similarly,

\displaystyle\longrightarrow\sf{\underline{\underline{x_2=\dfrac{2m_1\,x_0}{m_1+m_2}}}}⟶

x

2

=

m

1

+m

2

2m

1

x

0