Kvant Physics Problem 2
Two balls of masses $m_1$ and $m_2$ are placed on a smooth horizontal plane and connected by a spring of stiffness $c$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 7m53s
Source on kvant.digital
Problem
On a horizontal plane lie two balls of masses $m_1$ and $m_2$, connected to each other by a spring with stiffness $c$. The plane is smooth. The balls are displaced, compressing the spring, and then released simultaneously. Determine the periods of the resulting oscillations of the balls.
Setup and Assumptions
Two balls of masses $m_1$ and $m_2$ are placed on a smooth horizontal plane and connected by a spring of stiffness $c$. The spring is initially compressed and then both balls are released simultaneously.
The required quantity is the period of the oscillatory motion of each ball.
The given quantities are:
$$m_1 ; (\mathrm{kg}), \qquad m_2 ; (\mathrm{kg}), \qquad c ; (\mathrm{N/m}).$$
The following idealizations are adopted. The plane is perfectly smooth, so there is no friction. The spring is ideal, massless, and obeys Hooke's law. The oscillations are small enough that the spring force is proportional to the deformation. External horizontal forces are absent, so the center of mass moves with constant velocity. The motion is considered in an inertial reference frame fixed to the plane.
Since the balls are released simultaneously from rest, the total horizontal momentum is initially zero and remains zero. Consequently, the center of mass remains at rest.
Physical Principles
The solution is based on Newton's second law and Hooke's law.
If the spring is stretched or compressed by an amount $x$ relative to its natural length, the spring force magnitude is
$$F = cx.$$
For each mass, Newton's second law gives
$$m a = F.$$
Because the only horizontal forces are the spring forces, the internal forces on the two masses are equal in magnitude and opposite in direction.
It is convenient to describe the motion through the relative coordinate
$$r = x_2 - x_1,$$
where $x_1$ and $x_2$ are the coordinates of the masses along the line joining them.
Simple harmonic motion with angular frequency $\omega$ satisfies
$$\ddot q + \omega^2 q = 0,$$
and its period is
$$T = \frac{2\pi}{\omega}.$$
Derivation
Let $l_0$ be the natural length of the spring. The deformation of the spring is
$$\xi = r - l_0.$$
When $\xi>0$ the spring is stretched, and when $\xi<0$ it is compressed.
The spring exerts on mass $m_1$ the force
$$F_1 = c\xi,$$
directed toward mass $m_2$. On mass $m_2$ the force is
$$F_2 = -c\xi.$$
Newton's second law gives
$$m_1\ddot x_1 = c\xi,$$
$$m_2\ddot x_2 = -c\xi.$$
Subtracting the first equation from the second yields
$$\ddot r = \ddot x_2-\ddot x_1 = -\frac{c}{m_2}\xi-\frac{c}{m_1}\xi.$$
Since $\xi=r-l_0$,
$$\ddot r = -c\left(\frac1{m_1}+\frac1{m_2}\right)(r-l_0).$$
Introducing the relative displacement
$$q=r-l_0,$$
we obtain
$$\ddot q + c\left(\frac1{m_1}+\frac1{m_2}\right)q = 0.$$
This is the equation of simple harmonic motion. Comparing it with
$$\ddot q+\omega^2 q=0,$$
gives
$$\omega^2 = c\left(\frac1{m_1}+\frac1{m_2}\right).$$
Hence
$$\omega = \sqrt{c\left(\frac1{m_1}+\frac1{m_2}\right)}.$$
The period of the relative oscillation is
$$T = \frac{2\pi}{\omega} = \frac{2\pi} {\sqrt{c\left(\frac1{m_1}+\frac1{m_2}\right)}}.$$
Multiplying numerator and denominator by $\sqrt{m_1m_2}$ gives the equivalent form
$$T = 2\pi \sqrt{\frac{m_1m_2}{c(m_1+m_2)}}.$$
Since the center of mass is at rest, both balls execute harmonic oscillations with this same period.
Result
The oscillations of the two balls have angular frequency
$$\boxed{ \omega = \sqrt{c\left(\frac1{m_1}+\frac1{m_2}\right)} }.$$
Their common period is
$$\boxed{ T = \frac{2\pi} {\sqrt{c\left(\frac1{m_1}+\frac1{m_2}\right)}} = 2\pi \sqrt{\frac{m_1m_2}{c(m_1+m_2)}}. }$$
No numerical values for $m_1$, $m_2$, and $c$ are provided in the statement, so a numerical substitution cannot be carried out. The answer remains in symbolic form.
Sanity Checks
The quantity
$$\frac{m_1m_2}{m_1+m_2}$$
has units of kilograms. Dividing by $c$ gives
$$\frac{\mathrm{kg}}{\mathrm{N/m}} = \frac{\mathrm{kg}}{\mathrm{kg/s^2}} = \mathrm{s^2},$$
so the square root has units of seconds, which is the correct unit for a period.
If one mass becomes extremely large, for example $m_2\to\infty$, then
$$\frac{m_1m_2}{m_1+m_2} \to m_1,$$
and
$$T \to 2\pi\sqrt{\frac{m_1}{c}}.$$
This is exactly the period of a single mass $m_1$ attached to a spring whose other end is effectively fixed.
If the masses are equal,
$$m_1=m_2=m,$$
then
$$T = 2\pi\sqrt{\frac{m}{2c}}.$$
The effective inertia is reduced because both masses move symmetrically.
The step most vulnerable to error is the derivation of the relative-coordinate equation. A sign mistake in the subtraction of the two Newton equations would produce
$$\ddot q-\omega^2 q=0,$$
which describes exponential growth rather than oscillatory motion and would lead to an unphysical result.