Kvant Physics Problem 9

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Problem

On a horizontal table there is a small mass attached to the table by a long spring. Initially the spring was unstretched. Then the mass was displaced by 20 cm from the equilibrium position and released. The mass began oscillating along the spring. Due to friction, the amplitude of its oscillations decreases by 7% per period. How many oscillations will the mass perform before stopping? At what distance from the equilibrium position will it stop?

All-Union Correspondence Physics and Mathematics Olympiad (1967)

Setup and Assumptions

A small mass $m$ is attached to a fixed point on a horizontal table by a spring of stiffness $k$. Motion is one-dimensional along the spring axis. The equilibrium position corresponds to zero spring deformation.

At $t=0$ the mass is displaced to amplitude $A_0 = 20,\text{cm} = 0.20,\text{m}$ and released from rest, so the motion is oscillatory about equilibrium.

Dissipation is modeled phenomenologically by a constant fractional decrease of amplitude by $7%$ per period. The motion is assumed to remain approximately periodic, so the amplitude after each full oscillation is multiplied by a constant factor $q = 0.93$.

Air resistance is neglected except insofar as it is encoded in the given decay law. The table is taken as inertial.

The quantity required is the number of oscillations until the motion ceases and the final position of the mass.

Physical Principles

The amplitude evolution over successive periods is described by a geometric progression

$A_n = A_0 q^n,$

where $q = 0.93$.

The oscillation ceases only when the amplitude becomes zero, meaning the turning points coincide with the equilibrium position and no further mechanical energy remains in the oscillatory degree of freedom.

The displacement at complete rest corresponds to the equilibrium position of the spring, determined by the condition of zero restoring force.

Derivation

After $n$ completed oscillations the amplitude is

$A_n = 0.20 \cdot (0.93)^n \ \text{m}.$

The amplitude is strictly positive for every finite $n$ because $0 < 0.93 < 1$. The limit of the amplitude sequence is

$\lim_{n \to \infty} A_n = 0.$

Since the motion consists of successive oscillations with decreasing amplitude but no finite index $n$ satisfying $A_n = 0$, the number of oscillations required for complete cessation is not finite.

The mass approaches equilibrium asymptotically, so the limiting rest position is given by the equilibrium coordinate

$x_{\text{eq}} = 0.$

Result

The number of oscillations is unbounded,

$N = \infty.$

The final position is the equilibrium position,

$x_{\text{final}} = 0.20 \cdot (0.93)^\infty = 0,\text{m}.$

$$\boxed{N = \infty,\quad x_{\text{final}} = 0,\text{m}}$$

Sanity Checks

The factor $0.93$ is dimensionless, consistent with a fractional amplitude decay per period. The expression $A_n = A_0 q^n$ preserves units of meters for all $n$.

A finite stopping time would require either a threshold amplitude or a non-multiplicative loss mechanism such as dry friction producing a constant energy loss per cycle; neither is specified. Under a multiplicative decay law, the amplitude approaches zero only in the limit of infinite oscillation count, matching the derived result.

The final position coincides with the equilibrium point, consistent with the restoring force $F = -kx$ vanishing at $x=0$. A sign error in the decay factor would appear if $q>1$, which would lead to divergence rather than decay.