A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.8 cm. (Assume the position of the object is at the origin at t-o.) (a) Calculate the maximum value of its speed. cm/s (b) Calculate the maximum value of its acceleration. cm/s2 (c) Calculate the value of its speed when the object is 10.80 cm from the equilibrium position. cm/s (d) Calculate the value of its acceleration when the object is 10.80 cm from the equilibrium position cm/s2 (e) Calculate the time interval required for the object to move from x = 0 to x = 4.80 cm.

Explanation / Answer

a) w = sqrt [k / m] = sqrt [8 / 0.520] = 3.92

maximum speed = A w = 12.8 * 3.92

= 50.2 cm/s

b) maximum acceleration = A w2 = 12.8 * 3.922

= 196.7 cm/s2

c) KE = [1/2 * 8.00 * 0.1282]- [1/2 * 8.00 * 0.1082] = 0.01888

0.01888 = 1/2 * 0.520 * v2

v = 26.95 cm/s

d) a = k x / m = 8.00 * 0.108 / 0.520

= 166 cm/s2

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