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 10.2 cm. (Assume the position of the object is at the origin at t= 0.) (a) Calculate the maximum value of its speed. cm/s (b) Calculate the maximum value of its acceleration. cm/s^2 (c) Calculate the value of its speed when the object is 6.20 cm from the equilibrium position. cm/s (d) Calculate the value of its acceleration when the object is 6.20 cm from the equilibrium position. cm/s^2 (e) Calculate the time interval required for the object to move from x = 0 to x = 2.20 cm. s

Explanation / Answer

a) Solving for maximum speed:

v = [{(Xo)^2 - (X^2)}k/m]
v = [{(0.102)^2 - (0)^2}(8/0.52)] m/s


b)

Solving for maximum acceleration:
a = - (k/m)Xo
a = - (8/0.52)(0.102) m/s2


c) v = ?, when X = 0.062 m
v = [{(Xo)^2 - (X^2)}k/m]
v = [{(0.102)^2 - (0.062^2)}8/0.520] m/s


d)

a = ?, when X = 0.0602 m
a = - (k/m)X
a = - (8/0.520)(0.0602) m/s2


e) T = time interval for object to move from x = 0 to x = 2.2cm
a = -(k/m)X
a = -(8/0.52)(0.022) m/s2


T = [-(4^2)X/a]
T = [-(4^2)0.0700/-1.24
T = 1.49 sec ANSWER

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