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 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)=20coswt.calculate the position of the mass at any time , measured from equilibrium position and the angular frequency for which resonance occurs.
Simran Bhatia , 10 Years ago
Grade 11
