Question

A 4-kg mass stretches a spring 0.392 m. If the mass is released from 1 m below the equilibrium position with a downward velocity of 10 m/sec, what is the maximum displacement?

I have no idea how to solve this.

Answer 1

"A 4-kg mass stretch a spring 0.392m If the mass is related from 1m below the equilibrium position with a downward velocity of 10m/sec, what is the maximum displacement? Given that mass (m) = 4kg, elongation (e) = 0.392m Then force F = mg = 4.(9.8) = 39.2 N Now we find spring constant k, k = F/e, where e is elongation k = 39.2/0.392 k = 100 N/m The differential equation that models this spring-mass system is 4 x"" + 100 x = 0 or x"" + 25x = 0 Initial Conditions: - Initial Position: - x(0) = 1 Intial velocity: - x�(0) = 10 We solve the Initial value problem(Iv_p) x"" + 25x = 0, x(0) = 1, x�(0) = 10 Auxiliary equation is m^2 + 25 = 0 Right double arrow m = plusminus 5: Therefore General sloution of (1) is \r\n x(t) = -c_1 cos 5t + c_2 sin 5t (2) On applying initial conditions, we get x(0) = \r\n "