A weight of mass 1 kg attached to a spring with a force constant of 20 N/m is able to oscillate on a horizontal

steel rod Fig. The initial displacement from the position of equilibrium is 30 cm. Find how many swings the weight will make before stopping completely. One swing is the movement from maximum displacement to the equilibrium position (or back). For numerical calculation put `g=10m//s^(2)` and coefficient of friction `mu=0.05`.

Answer

11 swings, i.e. almost 3 complete oscillations. [Hint. Do the calculations using the law of conservation of energy in the presence of external friction.]

Explanation

The friction force F_("fr")=mumg` is independent of the velocity. and this makes it possible to discuss the problem from the standpoint of energy. Suppose the weight goes over from the initial state of maximum deflection from the position of equilibrium, characterized by the amplitude `A_(0)` to another, similar state with amplitude `A_(1)`. Then according to the law of conservation of energy,
`(1)/(2)kA_(0)^(2)-mumgA_(0)=(1)/(2)kA_(1)^(2)+mumgA_(1)`
from which it follows that
`A_(1)=A_(0)-(2mumg)/(k)`
The same will be true for all the subsequent oscillations, i.e. the amplitudes will form an arithmetical progression:
`A_(n)=A_(0)-(2mumg)/(k)`
The pendulum will stop when its amplitude becomes zero. Substituting `A_(n)=0` given `n=(kA_(0))/(2mumg)` Since all the amplitudes except the initial one are passed through twice the number of swings is
`N=2n-1=(kA_(0))/(mumg)-1`
As may be seen, due to friction the mechanical energy rather quickly transforms into internal energy, and the oscillations cease.