An object of mass 0.5 kg is attached to a spring and is performing simple harmon

Question

An object of mass 0.5 kg is attached to a spring and is performing simple harmonic motion in horizontal direction. A force of 40 N pulls the object through a distance of 0.1 m from the mean position. Find the following;
(a) The angular frequency of the oscillation.
(b) The maximum velocity of the object and the velocity at the given position (0.1 m from the mean position) if the amplitude of the motion is 0.4 m
(c) The kinetic energy at a distance of 0.1 m from the mean position.
(d) The elastic potential energy at a distance of 0.1 m from the mean position.

Explanation / Answer

(a) F=-kx when F= 40N, x=0.1m F=-kx -40N=-k(0.1) (the sign of force and displacement doesn't matter as soon as they are the opposite ) k=400N/m ?=v(k/m) ?=v(400/0.5) ?=28.2842rad/s (b) Vmax= Aw because V= -Aw sin (?t+f), so Aw will be the max voceity. Aw is the amplitude of the voceity curve. Vmax= 0.4(28.2842) Vmax=11.3137m/s (c) Total energy = KE + PE(spring/elastic PE) 1/2 (400)(0.4)^2 = KE + 1/2 (400)(0.1)^2 KE= 30 J (d) Elastic energy = = 1/2 (400)(0.1)^2 = 2J

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