Jean asked in Science & MathematicsPhysics · 5 years ago

Oscillating circle?

A mass m = 2.97 kg is attached to a spring of force constant k = 42.9 N/m and set into oscillation on a horizontal frictionless surface by stretching it an amount A = 0.13 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 ω, constant tangential speed V = ωA, and centripetal acceleration of constant magnitude ac = ω2A.

http://www.webassign.net/webassignalgphys1/16-p-04...

(a) magnitude of the maximum force experienced by the oscillating mass

(b) maximum kinetic energy of the oscillating mass

(c) total energy of the oscillating mass-spring system

(d) If the record of time starts when x = +A and v = 0, determine expressions for the displacement, velocity, and acceleration of the oscillating mass along the x axis at any time t later. (Your expression should be in terms of the variable t and other numerical values. Assume any numerical values in your expression are in standard SI units, but do not enter units into your expression.

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  • 5 years ago
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    a) F_max = k*x_max = k*A

    b) KE_max = SPE_max = 1/2*k*x_max^2 = 1/2*k*A^2

    c) Same as b)

    d) Use the components in the circle. The angle is w*t

    x = A*cos(w*t)

    v = -w*A*sin(w*t)

    a = -w^2*A*cos(w*t)

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  • 5 years ago

    Because x(0) = A = 0.13, x(t) = A*cosωt => V(t) = dx/dt = -ω*A*sinωt

    => Vmax = 3.8*0.13 = 0.494m/s

    We know ω =√k/m] = 3.8 rad/s

    F = kx so Fmax = kA = 42.9*0.13 = 5.577 N

    KEmax = ½mVmax² = ½*2.97*4.94² = 0.3625 J

    Total Energy = ½mV² + ½kx² = ½kx² at t = 0 = ½*42.9*0.13² = 0.3625 J

    x(t) = A*cosωt = 0.13*cos3.8t meters

    V(t) = dx/dt = -ω*A*sinωt = -0.494*sin3.8t m/s

    a(t) = dV/dt = -ω²*A*cosωt = -1.878*cos3.8t m/s²

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