A particle with mass 2.47 kg oscillates horizontally at the end of a horizontal

Question

A particle with mass 2.47 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.857 m and a duration of 125 s for 8.0 × 101 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 34.1% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position.

Explanation / Answer

P = 125 / 80 = 1.56 sec period f = 1/P = .64 / sec frequency w = 2 * pi * f = 4.02 / sec angular frequency w = (k / m)^1/2 k = w^2 m = 4.02^2 * 2.47 = 39.9 N/m spring constant Umax = 1/2 k A^2 = 1/2 * 39.9 * .857^2 = 14.7 J total potential energy 1/2 m V^2 = Umax where V is the maximut speed V = (2 * U / m)^1/2 = 3.45 m/s x = .341 A = .292 m 34.1 % of amplitude U = 1/2 k x^2 solve for U to get potential energy at .341 A KE = Umax - U solve for KE Solve for v using KE as above

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