A mass m = 2.87 kg is attached to a spring of force constant k = 38.9 N/m and se
ID: 2255523 • Letter: A
Question
A mass m = 2.87 kg is attached to a spring of force constant k = 38.9 N/m and set into oscillation on a horizontal frictionless surface by stretching it an amount A = 0.11 m from its equilibrium position and then releasing it. The figure below shows the oscillating mass and the particle on the associated reference circle at some time after its release. The reference circle has a radius A, and the particle traveling on the reference circle has a constant counterclockwise angular speed ?, constant tangential speed V = ?A, and centripetal acceleration of constant magnitude ac = ?2A.
x =?v =?
a =?
A mass m = 2.87 kg is attached to a spring of force constant k = 38.9 N/m and set into oscillation on a horizontal frictionless surface by stretching it an amount A = 0.11 m from its equilibrium position and then releasing it. The figure below shows the oscillating mass and the particle on the associated reference circle at some time after its release. The reference circle has a radius A, and the particle traveling on the reference circle has a constant counterclockwise angular speed ?, constant tangential speed V = ?A, and centripetal acceleration of constant magnitude ac = ?2A.
Explanation / Answer
1) W = sqrt(K/m) = sqrt(38.9/2.87)= 3.682 rad/s
a) V max = A*W = 404.973e?3 m/s
b) a_max= W^2*A = 1.491 m/s^2
c) F = k*A = 4.279 N
d) KEmax = 0.5*m*Vmax^2 = 235.345 J
e) PEmax = 0.5*K*A^2 = 0.5*38.9*0.11^2 = 235.345 J
f) TE = PE max = KE max = 235.345 J
2) X = A=0.11*cos(3.68t) ................
V = -404.973e?3*sin(3.68t).........
a = -1.491cos(3.68t) = ..............
3) X = 0.11*sin(3.68t)....................
V = 404.973e?3*cos(3.68t)...................
a = -1.491*sin(3.68t)
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