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a spring is stretched 5*10^2m by a force of 5*10^-4N.a mass of 0.001kg is placed on the lower end of the spring.after equilibri...Asked by swapnali
A spring is stretched 5 × 10−2 m by a force of 5 × 10−4 N. A mass of 0.01 kg is placed on the
lower end of the spring. After equilibrium has been reached, the upper end of the spring is
moved up and down so that the external force acting on the mass is given by F(t) = 20 cos wt.
Calculate (i) the position of the mass at any time, measured form the equilibrium position and
(ii) the angular frequency for which resonance occurs
lower end of the spring. After equilibrium has been reached, the upper end of the spring is
moved up and down so that the external force acting on the mass is given by F(t) = 20 cos wt.
Calculate (i) the position of the mass at any time, measured form the equilibrium position and
(ii) the angular frequency for which resonance occurs
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Answered by
Pankaj
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