Two particles of masses m 1 ​ and m 2 ​ are joined by a massless spring of natural length L and force constant k. Initially, m 2 ​ is resting on a table and I am holding m 1 ​ vertically above m 2 ​ at a height L. At time t=0, I project m 1 ​ vertically upward with initial velocity v 0 ​ . Find the positions of the two masses at any subsequent time t (before either mass returns to the table) and describe the motion

1. Setting up the coordinates

Take vertical upward direction as positive.

• $y_1(t)$ – height of mass $m_1$ (top one)
• $y_2(t)$ – height of mass $m_2$ (bottom one)
• Natural spring length $L$ (so, at rest initially the spring is neither stretched nor compressed).
• Spring extension $x(t)=y_1(t)-y_2(t)-L$

2. Forces on each mass

Because the spring is mass-less and vertical, Newton’s second law gives

$m_1 \ddot{y}_1 = -m_1 g - k\,x$ $m_2 \ddot{y}_2 = -m_2 g + k\,x$ (the spring pulls one mass down and the other up with equal magnitude $k x$).

3. Split the motion into two easy parts

This is the classic trick:

1. Centre of mass (COM) coordinate $Y_{cm}=\frac{m_1 y_1+m_2 y_2}{m_1+m_2}$ External force is only gravity $-(m_1+m_2)g$, so $\ddot{Y}_{cm}=-g.$ Solution:
   $Y_{cm}(t)=Y_{cm}(0)+V_{cm}(0)t-\frac{1}{2}gt^2.$

2. Relative (internal) coordinate $x(t)$ Using the reduced mass $\mu=\dfrac{m_1 m_2}{m_1+m_2}$ we get $\mu \ddot{x}+k x =0.$ This is simple harmonic motion with angular frequency $\omega = \sqrt{\frac{k}{\mu}}.$

4. Initial conditions (just after the throw, $t=0$)

• $y_1(0)=L$, $y_2(0)=0$ (spring natural length)
• $\dot{y}_1(0)=v_0$, $\dot{y}_2(0)=0$ (only the top mass is kicked)

From these we get

• $x(0)=0$
• $\dot{x}(0)=v_0$ (because $\dot{x}=\dot{y}_1-\dot{y}_2$)
• $Y_{cm}(0)=\dfrac{m_1 L}{m_1+m_2}$
• $V_{cm}(0)=\dfrac{m_1 v_0}{m_1+m_2}$

5. Solve the two parts

Relative motion

$x(t)=\frac{v_0}{\omega}\sin(\omega t).$

Centre-of-mass motion

$Y_{cm}(t)=\frac{m_1 L}{m_1+m_2}+\frac{m_1 v_0}{m_1+m_2}\,t-\frac{1}{2}g t^2.$

6. Convert back to individual positions

Because $y_1=y_2+L+x$ and both masses share the COM,

$y_1(t)=Y_{cm}(t)+\frac{m_2}{m_1+m_2}\left[L+x(t)\right]$ $y_2(t)=Y_{cm}(t)-\frac{m_1}{m_1+m_2}\left[L+x(t)\right].$

Plugging $x(t)$ gives the full answers.

7. Nature of the motion

• Whole system rises, slows under gravity, then falls – exactly like one body of mass $m_1+m_2$.
• Internally the two masses oscillate about their natural separation with amplitude $\dfrac{v_0}{\omega}$ and frequency $\omega$.

The expressions stay valid until the lower mass $y_2(t)$ touches the table again $(y_2=0)$.

Imagine two friends on a pogo-stick rope

1. Rope (spring) wants to keep its own happy length.
2. Bottom friend is standing on the floor (table).
3. Top friend is exactly one rope-length above the bottom friend – so the rope is not stretched or squashed.
4. Suddenly you throw the top friend straight upward.
   • He goes up and pulls the rope.
   • The rope pulls the bottom friend upward too.
5. Both friends now leave the floor and start a funny dance:
   • Together they feel Earth’s pull, so their centre follows an ordinary up-and-down parabola just like one stone.
   • Between themselves they stretch and shrink the rope in a smooth back-and-forth motion (like a spring toy).

Put both stories together and you get the complete motion of each friend!

Step-by-step mathematical solution

Let upward be positive.

1. Define variables

   y_1(t) &\;=\; \text{height of } m_1 \\ y_2(t) &\;=\; \text{height of } m_2 \\ x(t) &\;=\; y_1-y_2-L \quad (\text{spring extension}) \end{aligned}$$

2. Write Newton’s equations $\boxed{m_1 \ddot{y}_1 = -m_1 g - k x}$ $\boxed{m_2 \ddot{y}_2 = -m_2 g + k x}$

3. Split into COM and relative coordinates COM mass $M=m_1+m_2$, reduced mass $\mu=\dfrac{m_1 m_2}{M}$.

   $Y_{cm}=\frac{m_1 y_1+m_2 y_2}{M},\qquad \ddot{Y}_{cm}=-g.$ $\mu \ddot{x}+k x=0 \;\Rightarrow\; \ddot{x}+\omega^2 x=0,\quad \omega=\sqrt{\frac{k}{\mu}}.$

4. Initial conditions at $t=0$ $y_1(0)=L,\; y_2(0)=0,\; \dot{y}_1(0)=v_0,\; \dot{y}_2(0)=0$ Hence $x(0)=0,\; \dot{x}(0)=v_0$ $Y_{cm}(0)=\frac{m_1 L}{M},\; V_{cm}(0)=\frac{m_1 v_0}{M}.$

5. Solve relative (SHM) equation

   $x(t)=\frac{v_0}{\omega}\sin(\omega t).$

6. Solve COM equation

   $Y_{cm}(t)=\frac{m_1 L}{M}+\frac{m_1 v_0}{M}t-\frac{1}{2}g t^2.$

7. Recover individual positions Use $y_1=y_2+L+x$ and COM definition.

   y_1(t)&=Y_{cm}(t)+\frac{m_2}{M}\bigl[L+x(t)\bigr]\\[6pt] y_2(t)&=Y_{cm}(t)-\frac{m_1}{M}\bigl[L+x(t)\bigr] \end{aligned}}$$ Substitute $x(t)$: $$\boxed{\begin{aligned} y_1(t)&=\frac{m_1 L}{M}+\frac{m_1 v_0}{M}t-\frac{1}{2}g t^2+\frac{m_2}{M}\left[L+\frac{v_0}{\omega}\sin(\omega t)\right]\\[6pt] y_2(t)&=\frac{m_1 L}{M}+\frac{m_1 v_0}{M}t-\frac{1}{2}g t^2-\frac{m_1}{M}\left[L+\frac{v_0}{\omega}\sin(\omega t)\right] \end{aligned}}$$

8. Validity condition (before table contact) Solution holds while $y_2(t)\ge 0$.

9. Motion description

   • Centre of mass performs free-fall under gravity.
   • About that falling frame, the two masses execute simple harmonic oscillation of amplitude $\dfrac{v_0}{\omega}$ and frequency $\omega=\sqrt{\dfrac{k(m_1+m_2)}{m_1 m_2}}$.

   This combined motion continues until either mass hits the table again.