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A weight of mass 1 kg attached to a spring with a force constant of 20 N/m is able tooscillate 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. |
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Answer» `(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 numberof 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. |
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