A 0.223-kg particle undergoes simple harmonic motion along the horizontal x-axis

Question

A 0.223-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.255 m and x2 = 0.425 m. The period of oscillation is 0.601 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 of position=_________m

amplitude=__________m

maximum speed=_________m/s

max. mag. of accelerations=____________m/s^2

force constant=_________N/m

total mechanical energy=__________J

Explanation / Answer

a)Period = 0.601s
frequency f = 1 / T = 1.663 Hz

b) first find the amplitude , total displacement ,0.425 m+0.255=0.68m

so amplitude = 0.68m / 2 = 0.34 m

equilibrium postion =0.425m - 0.34m = +0.085 m

c) now, omega = 2f = 2*1.663 rad/s = 10.448 rad/s
so, max speed = A*omega = 0.34m * 10.448 = 3.55 m/s

d) maximum acceleration = A*omega2 = 0.34*10.4482 m/s2 = 37.11m/s2

e) Max Force = m*a

= 0.223kg * 37.11 m/s2

= 8.276 N
f) F = kA
so force constant k =

F / A = 8.276 N/ 0.34m

= 24.34 N/m

g) total energy = 0.5*m*v2 = 0.5*0.223*3.552 J = 1.405 J

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