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 8.90 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.)

? = rad/s f = Hz T = s A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.90 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.) What is the force constant of the spring?

Explanation / Answer

x = A cos w t since we don't need a phase angle here

w t = 26.9 * .5 = 13.45 rad

4 * pi = 12.57rad 2 complete oscillations

(13.45 - 12.6) * 360 / (2 * pi) = 48.7 deg (2 cycles + 48.7 deg)

A = .05 cos w t = .05 * cos 48.7 = .033 cm

v = -w A sin w t = -26.9 * .05 * sin 48.7 = -1.01 m/s

a = - w^2 A cos w t = -(26.9)^2 * .05 * cos 48.7 = -23.9 m/s^2

velocity is towards origin (negative)

acceleration is negative (decreasing from its maximum)

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