A 0.420-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.420-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 13.0 cm. (Assume the position of the object is at the origin at t = 0.) (a) Calculate the maximum value of its speed. 56.73665 Correct: Your answer is correct. cm/s (b) Calculate the maximum value of its acceleration. 247.6190476 Correct: Your answer is correct. cm/s2 (c) Calculate the value of its speed when the object is 11.00 cm from the equilibrium position. 131.9657758 Incorrect: Your answer is incorrect. The speed at this position cannot be greater than the maximum speed. cm/s (d) Calculate the value of its acceleration when the object is 11.00 cm from the equilibrium position. 209.5238095 Incorrect: Your answer is incorrect. Think carefully about the direction of the acceleration if the displacement is positive. cm/s2 (e) Calculate the time interval required for the object to move from x = 0 to x = 7.00 cm. s

Explanation / Answer

a) Solving for maximum speed:

v = [{(Xo)^2 - (X^2)}k/m]
v = [{(0.13)^2 - (0)^2}(8/0.420)]
v = 0.567 m/sec ANSWER

Solving for maximum acceleration:
a = - (k/m)Xo
a = - (8/0.420)(0.13)
a = - 2.47 m/sec^2

b) v = ?, when X = 0.1100 m
v = [{(Xo)^2 - (X^2)}k/m]
v = [{(0.13)^2 - (0.1100^2)}8/0.420]
v = 0.302 m/sec ANSWER

a = ?, when X = 0.1100 m
a = - (k/m)X
a = - (8/0.420)(0.1100)
a = - 2.09 m/sec^2 ANSWER

c) T = time interval for object to move from x = 0 to x = 7.00cm
a = -(k/m)X
a = -(8/0.42)(0.0700)
a = - 1.33 m/sec^2

T = [-(4^2)X/a]
T = [-(4^2)0.0700/-1.33
T = 1.44 sec ANSWER


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