The period of motion of an object-spring system is T = 0.510 s when an object of

Question

The period of motion of an object-spring system is T = 0.510 s when an object of mass m = 254 g is attached to the spring. (a) Find the frequency of motion in hertz. (b) Find the force constant of the spring. (c) If the total energy of the oscillating motion is 0.267 J, find the amplitude of the oscillations. A "seconds" pendulum is one that goes through its equilibrium position once each second. (The period of the pendulum is 2.000 s.) The length of a at Tokyo and 0.9942 m at Cambridge, England. What is the ratio of the free fall accelerations at these two factions? (Give your answer to at least 4) Cambridge/Tokyo = 1.0015

Explanation / Answer

a)

frequency of the motion = 1/Time period

frequency of the motion = 1/0.510

frequency of the motion = 1.96 Hz

b)

force constant is k

T = 2pi * sqrt(m/k)

0.510 = 2pi * sqrt(0.254/k)

solving for k

k = 38.6 N/m

the spring constant is 38.6 N/m

c)

let the amplitude is A

total energy = 0.50 * k * A^2

0.267 = 0.50 * 38.6 * A^2

solving for A

A = 0.118 m

the amplitude of motion is 0.118 m

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