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An impulsive force gives an initial velocity of -1.0 ms^(-1)to the mass in the unstretched spring position. What is the amplitude of motion ? Give x as a function of time for the oscillating mass. Given m=3 kg, k= 1200 Nm^(-1)

Answer»


Solution :`a = v_(MAX)//omega =v_(max) = 1xxsqrt(3//120) m, omega = sqrt(k//m) = 20 "RAD s"^(-1) , X = - 0.05 COS 20t`.


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