An .86 kg mass at the end of a spring vibrates 3.0 times per second with an ampl

Question

An .86 kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of .13 m. Determine 1) the velocity when it is .10 m from equilibrium, 2) the equation describing the motion of the mass, assuming that x was a maximum of t=0, and 3) The velocity and acceleration functions for the mass.

Explanation / Answer

given n = 6 Hz, A = 0.13 m & m = 0.86 kg =>By n = 1/2pv[k/m] =>2pn = v[k/m] =>k = (2pn)^2 x m =>k = (2 x 3.14 x 6)^2 x 0.86 =>k = 1221.01 N/m =>PE(max) = 1/2kA^2 = 1/2 x1221.01 x (0.1)^2 = 10.31 J (a) By the law of energy conservation:- =>KE(max) = PE(max) =>1/2mv^2 = 10.31 =>v = v[(2 x 10.31)/0.86] =>v = 4.89 m/s (b) By the law of energy conservation:- =>PE(max) = PE(at0.08m) + KE(at0.08m) =>1/mv^2 = 10.31 - 1/2 x 1221.01 x (0.08)^2 =>v= (c) Total energy = PE(max) = 10.31J (d) By x(t) = Acos(?t + f)

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