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A 0.115-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.267 m and x2

Posted: Mon May 02, 2022 9:16 pm
by answerhappygod
A 0 115 Kg Particle Undergoes Simple Harmonic Motion Along The Horizontal X Axis Between The Points X1 0 267 M And X2 1
A 0 115 Kg Particle Undergoes Simple Harmonic Motion Along The Horizontal X Axis Between The Points X1 0 267 M And X2 1 (127.19 KiB) Viewed 23 times
A 0.115-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.267 m and x2 = 0.499 m. The period of oscillation is 0.543 s. Find the frequency, f, the equilibrium position, Xeq, the amplitude, A, the maximum speed, Umax, the maximum magnitude of acceleration, Amax, the force constant, k, and the total mechanical energy, Etot f = Hz Xeq = m A= m Umax = m/s amax = m/s2 k= N/m Etot = J