A 0.225-kg particle undergoes simple harmonic motion along the horizontal x-axis
ID: 2098194 • Letter: A
Question
A 0.225-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.259 m and x2 = 0.411 m. The period of oscillation is 0.655 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.
frequency=__HZ
equilibrium postition=____m
A=___m
max speed=___m/s
max magnitude=___m/s^2
k=___N/m
total mechanical energy=___J
Explanation / Answer
f= 1/T = 1/0.655 = 1.527 s^-1 equilibrium position = 0.411- [(0.411+0.259)/2] = 0.076 Amplitude = (0.411+0.259)/2 = 0.335 w= 2*pi * f= 9.59 so, K= w^2 *m = 20.693 1/2 mv^2 = 1/2 K * A^2 => v= 3.213 m/s Total mechanical energy = 1/2 K* A^2 = 1.16 J maxm. acceleration = w^2 .A = 30.809 m/s2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.