A spring of negligible mass stretches 3.00 cm from its relaxed length when a for

Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 6.00 N is applied. A 0.410-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (Assume that the direction of the initial displacement is positive.)

(a) What is the force constant of the spring?
N/m

(b) What are the angular frequency (?), the frequency, and the period of the motion?

Explanation / Answer

F=kx the angular frequency is: omega = sqrt (k/m) = sqrt (F / xm) Total energy at max extension, X, is PE = 1/2 kX^2 = 1/2 FX^2/x The amplitude is just X, the max extension To get the max velocity at equilibrium, use conservation of energy PE at max extension = 1/2 FX^2/x = KE at equilibrium = 1/2 mv^2 v = sqrt (FX^2 / (xm)) The displacement as a function of time is d(t) = X cos (omega t) = X cos (sqrt (F / xm) t) They give you the little extension (x),the force at that extension (F), the max extension (X), the mass (m), and the time for part f (t).

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