A particle with a mass of 0.500 kg is attached to a spring with aforce constant

Question

A particle with a mass of 0.500 kg is attached to a spring with aforce constant of 50.0 N/m. At the moment, t = 0, the particle hasits maximum speed of 20.0 m/s and is moving to the left.

a) Determine the particle’s equation of motion, specifyingits position as a function of time. Justify the phase constant thatyou used.

b) Where in the motion is the potential energy three times thekinetic energy?

c) Find the minimum time interval required for the particle to movefrom x = 0 to x = 1.00 m.

d) Find the length of a simple pendulum with the same period.

Explanation / Answer

m = 0.500 kg, k = 50.0 N/m, = (k/m) = 10 rad/s t = 0, v(0) = -20.0 m/s, x(0) = 0 a) x(t) = Asin(t + ), x(0) = Asin = 0 v(t) = Acos(t + ), v(0) = Acos =10Acos = -20 = , A = 2.0 m x(t) = 2.0 sin(10t + ) b) Ep = 3Ek Ek + Ep = E Ep/3 + Ep = E 4Ep/3 = E (4/3)kx2/2 = kA2/2 x = ±3A/2 c) at t, x = 0, x(t) = 2.0sin(10t + ) = 0, 10t + =0 at t', x = 1.00 m, x(t') = 2.0sin(10t' + ) = 1.0, 10t' + = /6 10(t' - t) = /6 t' - t = /60 = 0.0524 s d) period T = 2/ = /5 T = 2(L/g) = /5 L = g/100 = 9.8 cm

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