Kvant Physics Problem 9

A small mass moves on a horizontal table attached to a fixed point by a spring.

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Problem

A small mass is on a horizontal table, attached to the table by a long spring. Initially the spring is unstretched. The mass is then displaced by 20 cm from its equilibrium position and released. The mass begins to oscillate along the spring. Due to friction, the amplitude of its oscillations decreases by 7% each period. How many oscillations in total will the mass make before coming to rest? At what distance from the equilibrium position will it come to rest?

All-Union Correspondence Physics and Mathematics Olympiad (1967)

Setup and Assumptions

A small mass moves on a horizontal table attached to a fixed point by a spring. The equilibrium position corresponds to the unstretched spring. The mass is displaced initially by

$A_0 = 20\ \text{cm}$

and released. The problem states that after each complete period the amplitude decreases by $7%$. Let $A_n$ denote the amplitude after $n$ complete oscillations. The given information implies the recurrence relation

$A_{n+1} = 0.93, A_n.$

Hence the amplitudes form a geometric progression

$A_n = A_0 (0.93)^n.$

No assumption about the type of friction is needed; the fractional decrease per period is the essential datum.

Physical Principles

The relation $A_{n+1} = 0.93, A_n$ describes a repeated fractional decay of the amplitude. Repeated application gives

$A_n = A_0 (0.93)^n.$

Since $0 < 0.93 < 1$, the limit as $n \to \infty$ satisfies

$\lim_{n \to \infty} A_n = A_0 \lim_{n \to \infty} (0.93)^n = 0.$

Mathematically the amplitude decreases to zero only in the limit of infinitely many oscillations. Physically, the mass will stop oscillating once the amplitude becomes smaller than any practically measurable threshold. If the motion is considered negligible when the amplitude falls below $A_{\text{min}} = 0.1\ \text{cm}$, the total number of oscillations until the mass can be considered at rest can be estimated by solving

$A_n = A_0 (0.93)^n \le A_{\text{min}}.$

Taking the logarithm of both sides gives

$\ln (0.93^n) \le \ln \frac{A_{\text{min}}}{A_0},$

which simplifies to

$n \ln 0.93 \le \ln \frac{0.1}{20}.$

Since $\ln 0.93$ is negative, dividing both sides by $\ln 0.93$ reverses the inequality:

$n \ge \frac{\ln(0.1/20)}{\ln 0.93}.$

Computing each term explicitly, the numerator is

$\ln \frac{0.1}{20} = \ln 0.005 \approx -5.2983,$

and the denominator is

$\ln 0.93 \approx -0.07232.$

Thus the total number of oscillations until the amplitude falls below $0.1\ \text{cm}$ is

$n \approx \frac{-5.2983}{-0.07232} \approx 73.3.$

Since the number of complete oscillations must be an integer, the mass executes approximately $74$ oscillations before coming to rest in a practical sense.

Derivation

Starting from

$A_n = A_0 (0.93)^n$

with $A_0 = 20\ \text{cm}$, the amplitude after $n$ periods is

$A_n = 20 \cdot (0.93)^n\ \text{cm}.$

For each finite integer $n$, $A_n > 0$, so the motion continues across the equilibrium position while the amplitude gradually decreases. The amplitude falls below $0.1\ \text{cm}$ after approximately $74$ oscillations. Therefore the total number of oscillations before the mass can be considered at rest is

$N \approx 74.$

To determine the final position, note that the amplitude decreases symmetrically around the equilibrium position. The limiting displacement after the motion has practically ceased is

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

Answer

The mass executes approximately

$N \approx 74$

oscillations before coming to rest, and it comes to rest at the equilibrium position

$x_{\text{final}} = 0\ \text{cm}.$

The motion consists of oscillations of gradually decreasing amplitude, and the displacement at which it effectively stops coincides with the spring's equilibrium.